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Proof, EXT2 EQ-Bank 25

Use mathematical induction to prove De Moivre's theorem:

\((\cos \theta+i \,\sin \theta)^n=\cos n \theta+i \, \sin n \theta\)

for all integers  \(n \geq 1\).   (3 marks)

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\(\text{Proof (See Worked Solutions)}\)

Show Worked Solution

\(\text{Prove} \ \ (\cos \theta+i \, \sin \theta)^n=\cos (n \theta)+i \sin (n \theta) \ \text {for} \ \ n \geq 1\)

\(\text{If} \ \ n=1:\)

\((\cos \theta+i \, \sin \theta)^1=\cos \theta+i \, \sin \theta=\cos (1 \theta)+i \, \sin (1 \theta)\)

\(\therefore \ \text{True for} \ \ n=1\)
 

\(\text{Assume true for} \ \  n=k:\)

\((\cos \theta+i \, \sin \theta)^k=\cos k \theta+i \,\sin k \theta\ \ldots\ (1)\)

\(\text{Prove true for} \ \ n=k+1:\)

\(\text{i.e.}\ \ (\cos \theta+i \,\sin \theta)^{k+1}=\cos (k+1) \theta+i \, \sin (k+1) \theta.\)

\(\text{LHS}\) \(=(\cos \theta+i \, \sin \theta)^{k+1}\)
  \(=(\cos \theta+i \, \sin \theta)^k(\cos \theta+i \, \sin \theta)\)
  \(=(\cos k \theta+i \,  \sin k \theta)(\cos \theta+i \, \sin \theta) \ \ \text{(using}\ (1)\ \text{above)}\)
  \(=\cos k \theta \, \cos \theta+i \,  \cos k \theta \,\sin \theta+i \, \sin k \theta \,\cos \theta+i^2 \,  \sin k \theta \,\sin \theta\)
  \(=(\cos k \theta \, \cos \theta-\sin k \theta \,\sin \theta)+i(\sin k \theta \,\cos \theta+\cos k \theta \,\sin \theta)\)
  \(=\cos (k+1) \theta+i \, \sin (k+1) \theta\)

  

\(\Rightarrow \text{True for} \ \ n=k+1\)

\(\therefore \ \text{Since true for \(n=1\), by PMI, true for integers \(n \geqslant 1\).}\)

Proof, EXT2 EQ-Bank 34

Consider a sequence of rectangles with side lengths \(a_{ n }\) and \(b_{ n }\).

The first rectangle has  \(a_1=2\)  and  \(b_1=1\).

For integers  \(n \geq 1,\ \ a_{n+1}=\dfrac{a_n+b_n}{2}\)  and  \(b_{n+1}=\dfrac{2}{a_{n+1}}.\)

  1. Show that every rectangle in the sequence has an area of 2 square units.   (1 mark)

  2. Use the relationship between the arithmetic mean and the geometric mean to prove that  \(a_n \geq \sqrt{2}\)  for any integer  \(n \geq 1\).   (2 marks)

  3. Use mathematical induction to prove that  \(a_n-\sqrt{2} \leq \dfrac{1}{2^{n-1}}(2-\sqrt{2})\)  for any integer  \(n \geq 1\).    (4 marks)

  4. Use the squeeze theorem to show that the rectangles approach a square as \(n\) approaches infinity.   (2 marks)

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Show Worked Solution

a.    \(\text{Since}\ \ a_1 b_1=2\ \ \text{and}\ \ a_{n+1} b_{n+1}=a_{n+1} \times \dfrac{2}{a_{n+1}}=2\)

\(\Rightarrow\ \text{Each rectangle has area 2.}\)
 

b.    \(a_1=2\ \ \Rightarrow\ \ a_1 \geq \sqrt{2}\ \ \text{(true for 1st rectangle)}\)

\(\text{AM/GM inequality:}\ \ \dfrac{a_n+b_n}{2} \geq \sqrt{a_nb_n} \)

\(a_nb_n=2\ \ \text{(from part (a))}\)

\(\dfrac{a_n+b_n}{2} \geq \sqrt{2}\ …\ (1)\)

\(\text{Since}\ \ a_{n+1}=\dfrac{a_n+b_n}{2}:\)

\(\ a_{n+1} \geq \sqrt{2}\ \ \ \text{(using (1) above)}\)

\(\therefore a_n \geq \sqrt{2}\)
  

c.    \(\text{Prove}\ \ a_n-\sqrt{2} \leq \dfrac{1}{2^{n-1}}(2-\sqrt{2}),\ \ \text{for}\ \ n \geq 1\)

\(\text{If}\ \ n=1:\)

\(\ a_1-\sqrt{2}=2-\sqrt{2} \leq \dfrac{1}{2^{0}}(2-\sqrt{2})\).

\(\Rightarrow\ \text{True for}\ \ n=1.\)
 

\(\text{Assume true for}\ \ n=k:\)

\(a_k-\sqrt{2} \leq \dfrac{1}{2^{k-1}}(2-\sqrt{2})\ …\ (1) \)

\(\text{Prove true for}\ \ n=k+1:\)

\(\text{i.e.}\ \ a_{k+1}-\sqrt{2} \leq \dfrac{1}{2^k}(2-\sqrt{2})\)

\(\text{By definition,} \ \ a_{k+1}=\dfrac{1}{2}\left(a_k+b_k\right)\)

\(a_{k+1}-\sqrt{2}\) \(=\dfrac{1}{2}\left(a_k-\sqrt{2}\right)+\dfrac{1}{2}\left(b_k-\sqrt{2}\right) \)  
  \(\leq \dfrac{1}{2}\left(\dfrac{1}{2^{k-1}}(2-\sqrt{2})\right)+\dfrac{1}{2}\left(b_k-\sqrt{2}\right)\ \ \text{(see (1) above)}\ \)  

 
\(\text{Since}\ \ a_k \geq \sqrt{2}\ \ \text{and}\ \ a_kb_k=2:\)

\(\ b_k \leq \sqrt{2}\ \ \text{and}\ \ b_k-\sqrt{2} \leq 0\)

\(a_{k+1}-\sqrt{2} \leq \dfrac{1}{2^k}(2-\sqrt{2})+\dfrac{1}{2}\left(b_k-\sqrt{2}\right) \leq \dfrac{1}{2^k}(2-\sqrt{2})\)

\(\Rightarrow\ \text{True for}\ \ n=k+1.\)

\(\therefore\ \text{Since true for}\ \ n=1,\ \text{by PMI, true for integers}\ \ n \geq 1.\)
  

d.    \(\text {Combining parts (b) and (c):}\)

\(0 \leq a_n-\sqrt{2} \leq \dfrac{1}{2^{n-1}}(2-\sqrt{2}).\)

\(\text{As}\ \ n \rightarrow \infty, \ \dfrac{1}{2^{n-1}} \rightarrow 0\)

\(\Rightarrow a_n-\sqrt{2} \rightarrow 0\ \ \text{(by squeeze theorem)}\)

\(\text{Since rectangles have an area = 2:}\)

\(\text{As}\ \ n \rightarrow \infty,\ b_n \rightarrow \sqrt{2}\)

\(\text{i.e. rectangles approach a square.}\)

Proof, EXT2 P2 2025 HSC 12b

Given the function  \(y=x e^{2 x}\),  use mathematical induction to prove that  \(\dfrac{d^n y}{d x^n}=\left(2^n x+n 2^{n-1}\right) e^{2 x}\)  for all positive integers \(n\), where  \(\dfrac{d^n y}{d x^n}\) is the 
\(n\)th derivative of \(y\) and \(\dfrac{d}{d x}\left(\dfrac{d^n y}{d x^n}\right)=\dfrac{d^{n+1} y}{d x^{n+1}}\).   (3 marks)

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\(\text{Proof (See worked solutions)}\)

Show Worked Solution

\(\text{Prove} \ \ \dfrac{d^n y}{d x^n}=\left(2^n x+n 2^{n-1}\right) e^{2 x}\)

\(\text{If } \ \ n=1:\)

\(\operatorname{LHS}=\dfrac{d}{d x}\left(x e^{2 x}\right)=x \cdot 2 e^{2 x}+1 \cdot e^{2 x}=(2 x+1) e^{2 x}\)

\(\operatorname{RHS}=\left(2^{1} \cdot x+1 \cdot 2^{0}\right) e^{2 x}=(2 x+1) e^{2 x}=\operatorname{RHS}\)

\(\therefore \ \text{True for} \ \ n=1\)
 

\(\text{Assume true for}\ \  n=k:\)

\(\dfrac{d^k y}{d x^k}=\left(2^k x+k\, 2^{k-1}\right) e^{2 x}\)
 

\(\text{Prove true for}\ \  n=k+1:\)

\(\text{i.e.}\ \dfrac{d^{k+1} y}{d x^{k+1}}=\left(2^{k+1} x+(k+1) 2^k\right) e^{2 x}\)

\(\dfrac{d^{k+1} y}{d x^{k+1}}\) \(=\dfrac{d}{d x}\left(2^k x+k\cdot 2^{k-1}\right) e^{2 x}\)
  \(=\dfrac{d}{d x}\left(2^k x e^{2 x}+k\cdot 2^{k-1} e^{2 x}\right)\)
  \(=2^k x \cdot 2 e^{2 x}+2^k \cdot e^{2 x}+k \cdot 2^{k-1} \cdot 2 e^{2 x}\)
  \(=2^{k+1} x e^{2 x}+2^k \cdot e^{2 x}+k \cdot 2^k \cdot e^{2 x}\)
  \(=2^{k+1} x e^{2 x}+(k+1) 2^k \cdot e^{2 x}\)
  \(=\left(2^{k+1} x+(k+1) 2^k\right) e^{2 x}\)

 

\(\Rightarrow \ \text{True for} \ \ n=k+1\)

\(\therefore \ \text{Since true for \(\ \ n=1\), by PMI, true for integers} \ \ n \geqslant 1.\)

Proof, EXT2 P2 2024 HSC 14b

Use mathematical induction to prove that  \({ }^{2 n} C_n<2^{2 n-2}\),  for all integers  \(n \geq 5\).   (3 marks)

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\(\text{Prove  \({ }^{2 n} C_n<2^{2 n-2}\)  for  \(n \geqslant 5\).}\)

\(\text {If}\ \ n=5:\)

\(\text{LHS}={ }^{10} C_5=252\)

\(\text{RHS}=2^8=256>\text {LHS }\)

\(\therefore\ \text{True for}\ \ n=5\)
 

\(\text {Assume true for } n=k:\)

   \({ }^{2 k} C_k<2^{2 k-2}\ \ldots\ (1)\)

\(\text{Prove true for}\ \ n=k+1:\)

   \(\text{i.e. } {}^{2 k+2}C_{k+1}<2^{2 k}\)

  \(\text{LHS}\) \(=\dfrac{(2 k+2)!}{(k+1)!(k+1)!}\)
    \(=\dfrac{(2 k)!}{k!k!} \times \dfrac{(2 k+1)(2 k+2)}{(k+1)(k+1)}\)
    \(={ }^{2 k} C_k \times 2 \times \dfrac{2 k+1}{k+1}\)
    \(< 2^{2k-2} \times 2 \times \dfrac{2 k+1}{k+1}\)
    \(<2^{2k} \ \ \ \left(\text{since}\ \dfrac{2 k+1}{k+1}=2-\dfrac{1}{k+1}<2\right)\)

 
\(\Rightarrow \text{True for}\ \ n=k+1\)

\(\therefore \text{ Since true for \(n=5\), by PMI, true for integers \(n \geqslant 5\).}\)

Show Worked Solution

\(\text{Prove  \({ }^{2 n} C_n<2^{2 n-2}\)  for  \(n \geqslant 5\).}\)

\(\text {If}\ \ n=5:\)

\(\text{LHS}={ }^{10} C_5=252\)

\(\text{RHS}=2^8=256>\text {LHS }\)

\(\therefore\ \text{True for}\ \ n=5\)
 

\(\text {Assume true for } n=k:\)

   \({ }^{2 k} C_k<2^{2 k-2}\ \ldots\ (1)\)

\(\text{Prove true for}\ \ n=k+1:\)

   \(\text{i.e. } {}^{2 k+2}C_{k+1}<2^{2 k}\)

  \(\text{LHS}\) \(=\dfrac{(2 k+2)!}{(k+1)!(k+1)!}\)
    \(=\dfrac{(2 k)!}{k!k!} \times \dfrac{(2 k+1)(2 k+2)}{(k+1)(k+1)}\)
    \(={ }^{2 k} C_k \times 2 \times \dfrac{2 k+1}{k+1}\)
    \(< 2^{2k-2} \times 2 \times \dfrac{2 k+1}{k+1}\)
    \(<2^{2k} \ \ \ \left(\text{since}\ \dfrac{2 k+1}{k+1}=2-\dfrac{1}{k+1}<2\right)\)

 
\(\Rightarrow \text{True for}\ \ n=k+1\)

\(\therefore \text{ Since true for \(n=5\), by PMI, true for integers \(n \geqslant 5\).}\)

Proof, EXT2 P2 2023 SPEC1 8

A function \(f\) has the rule  \(f(x)=x\,e^{2x}\).

Use mathematical induction to prove that  \(f^{(n)}(x)=\big{(}2^{n}x+n\,2^{n-1}\big{)}e^{2x}\)  for  \(n \in \mathbb{Z}^{+}\), where  \(f^{(n)}(x)\)  represents the \(n\)th derivative of \(f(x)\). That is, \(f(x)\) has been differentiated \(n\) times.   (3 marks)

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\(\text{Proof (See Worked Solutions)}\)

Show Worked Solution

\(\text{If}\ \ n=1:\)

\(f(x)=x\,e^{2x}\ \ \Rightarrow \ \ f^{′}(x)=e^{2x} + 2x\,e^{2x} \)

\(f^{(1)}(x)=\big{(}2^{1}x + 1 \times 2^{1-1}\big{)}e^{2x} = e^{2x} + 2x\,e^{2x} \)

\(\therefore\ \text{True for}\ \ n=1.\)
 

\(\text{Assume true for}\ \ n=k \)

\(f^{(k)}(x)=\big{(}2^{k}x + k\,2^{k-1}\big{)}e^{2x}\)
 

\(\text{Prove true for}\ \ n=k+1 \)

\(\text{i.e.}\ f^{(k+1)}(x)=\big{(}2^{k+1}x + (k+1)2^{k}\big{)}e^{2x}\)

\(\text{LHS}\) \(= \dfrac{d}{dx} \big{(}f^{(k)}(x) \big{)} \)   
  \(= \dfrac{d}{dx} \Big{[}\big{(}2^{k}x + k\,2^{k-1}\big{)}e^{2x}\Big{]} \)  
  \(= 2^{k}e^{2x}+(2^{k}x+k\,2^{k-1}) \times 2 e^{2x} \)  
  \(=e^{2x} \big{(}2^{k}+2\cdot2^{k}x + 2k \cdot 2^{k-1}\big{)}\)  
  \(=e^{2x}\big{(}2^{k+1}+2^{k}+k \cdot 2^{k} \big{)} \)  
  \(=\big{(}2^{k+1}x + (k+1)2^{k}\big{)}e^{2x}\)  
  \(=\ \text{RHS}\)  

 
\(\Rightarrow \text{True for}\ \ n=k+1\)

\(\therefore\ \text{Since true for}\ \ n=1\text{, by PMI, true for}\ n \in \mathbb{Z}^{+} \)

Proof, EXT2 P2 2023 HSC 13b

  1. Show that  \(k^2-2 k-3 \geq 0\)  for  \(k \geq 3\).  (1 mark)

  2. Hence, or otherwise, use mathematical induction to prove that
  3.   \(2^n \geq n^2-2\), for all integers  \(n \geq 3\).  (3 marks)

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i.    \(\text{Proof (See Worked Solutions)} \)

ii.   \(\text{Proof (See Worked Solutions)} \)

Show Worked Solution

i.     \(k^2-2k-3\) \(=0\)
  \( (k-3)(k+1) \) \(=0\)

\(\text{Vertex (min) at}\ (1,-4) \)

\(\text{Quadratic is monotonically increasing for}\ \ k \geq1 \)

\(\text{At}\ \ k=3, \ k^2-2k-3=0 \)

\( \therefore\ \ k^2-2 k-3 \geq 0\ \ \text{for}\ \ k \geq 3\)
 

ii.    \(\text{Prove}\ \ 2^n \geq n^2-2 \)

\(\text{If}\ \ n=3: \)

\(\text{LHS}\ = 2^3=8 \)

\(\text{RHS}\ =3^3-2=7 \leq \text{LHS} \)

\(\therefore \ \text{True for}\ \ n=3. \)
 

\(\text{Assume true for}\ \ n=k: \)

\(2^k \geq k^2-2 \ \ \ …\ (*) \)
 

\(\text{Prove true for}\ \ n=k+1: \)

\(\text{i.e.}\ \ 2^{k+1} \geq (k+1)^2-2 \)

\(\text{LHS}\) \(=2^{k+1} \)  
  \(=2 \cdot 2^{k} \)  
  \( \geq 2(k^2-2) \ \ \ \text{(see (*) above)} \)  
  \( \geq 2k^2-4 \)  
  \( \geq k^2 + \underbrace{k^2-2k-3}_{\geq 0\ \  \text{(see part (i))}} +2k-1 \)  
  \(\geq k^2+2k-1 \)  
  \(\geq k^2+2k+1-2 \)  
  \(\geq (k+1)^2-2 \)  

 
\(\Rightarrow \text{True for}\ \ n=k+1 \)

\(\therefore \text{Since true for}\ \ n=3,\ \text{by PMI, true for integers}\ \ n\geq3 \)

Proof, EXT2 P2 EQ-Bank 11

A sequence is given by the recursive formula

`a_1=10, \ a_(n+1)=3a_n+4`  for  `n>=1`

Using mathematical induction to show the formula for the general term of the sequence is

`a_n=4(3^n)-2`   (3 marks)

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`text{Proof (See Worked Solution)}`

Show Worked Solution

`text{Prove true for}\ \ n=1:`

`text{LHS}\ =a_1=10`

`text{RHS}\ =4(3^1)-2=10=\ text{LHS}`

`:.\ text{True for}\ \ n=1`
 

`text{Assume true for}\ \ n=k:`

`a_k=4(3^k)-2\ \ text{… (1)}`
 

`text{Prove true for}\ \ n=k+1:`

`text{i.e.}\ a_(k+1)=4(3^(k+1))-2`

`a_(k+1)` `=3a_k+4`  
  `=3[4(3^k)-2]+4`  
  `=4*3^k*3-6+4`  
  `=4(3^(k+1))-2`  
  `=\ text{RHS}`  

 
`:.\ text{True for}\ \ n=k+1`

`:.\ text{Since true for} \ n=1,\ text{by PMI, true for integers} \ n>=1.`

Proof, EXT2 P2 EQ-Bank 18

Using mathematical induction, show

  `sum_(r=1)^n r^2=(n(n+1)(2n+1))/6`  for  `n>=1.`   (3 marks)

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`text{Proof (See Worked Solution)}`

Show Worked Solution

`text{Prove true for}\ \ n=1:`

`text{LHS}\ =sum_(r=1)^1 n^2=1^2=1`

`text{RHS}\ =(1xx2xx3)/6=1=\ text{LHS}`

`:.\ text{True for}\ \ n=1`
 

`text{Assume true for}\ \ n=k:`

`sum_(r=1)^k r^2=(k(k+1)(2k+1))/6`
 

`text{Prove true for}\ \ n=k+1:`

`text{i.e.}\ sum_(r=1)^(k+1) r^2=((k+1)(k+2)(2k+3))/6`

`sum_(r=1)^(k+1) r^2` `=sum_(r=1)^k r^2+(k+1)^2`  
  `=(k(k+1)(2k+1))/6+(k+1)^2`  
  `=((k+1)[2k^2+k+6k+6])/6`  
  `=((k+1)(2k^2+7k+6))/6`  
  `=((k+1)(k+2)(2k+3))/6`  
  `=\ text{RHS}`  

COMMENT: Required proof – preserve the `(k+1)` factor in all algebra!

`:.\ text{True for}\ \ n=k+1`

`:.\ text{Since true for} \ n=1,\ text{by PMI, true for integers} \ n>=1.`

Proof, EXT2 P2 EQ-Bank 16

Using mathematical induction, show

  `1/4n^4<sum_(r=1)^n r^3<=n^4`  for  `n>=1.`   (4 marks)

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`text{Proof (See Worked Solution)}`

Show Worked Solution

`text{Prove true for}\ \ n=1:`

`text{LHS}\ =1/4`

`text{Middle}\ =1^3=1`

`text{RHS}\ =1^4=1`

`1/4<1<=1`

`:.\ text{True for}\ \ n=1`
 

`text{Assume true for}\ \ n=k:`

`text{i.e.}\ 1/4k^4<sum_(r=1)^k r^3<=k^4`
 

`text{Prove true for}\ \ n=k+1:`

`text{i.e.}\ 1/4(k+1)^4<sum_(r=1)^(k+1) r^3<=(k+1)^4`

`text{LHS}\ =1/4(k+1)^4=1/4(k^4+4k^3+6k^2+4k+1)`

`text{RHS}\ =k^4+4k^3+6k^2+4k+1`

`text{Middle}\ =sum_(r=1)^(k+1) r^3=sum_(r=1)^k r^3+(k+1)^3`

`text{Consider LHS to show}\ \ 1/4(k+1)^4<sum_(r=1)^(k+1) r^3`

`text{Middle}` `=sum_(r=1)^k r^3+(k+1)^3`  
  `>1/4k^4+(k^3+3k^2+3k+1)`  
  `>1/4(k^4+4k^3+12k^2+12k+4)`  
  `>1/4(k^4+4k^3+6k^2+4k+1), \ \ (k>=1)`  
  `>\ text{LHS}`  

 

`text{Consider RHS to show}\ \ sum_(r=1)^(k+1) r^3<=k^4`

`text{Middle}` `=sum_(r=1)^k r^3+(k+1)^3`  
  `<=k^4+k^3+3k^2+3k+1`  
  `<=k^4+4k^3+6k^2+4k+4,\ \ (k>=1 => 4k^3>k^3, 6k^2>3k^2, 4k>3k)`  
  `<=(k+1)^4`  
  `<=\ text{RHS}`  

 

`:.\ text{True for}\ \ n=k+1`

`:.\ text{Since true for} \ n=1,\ text{by PMI, true for integers} \ n>=1.`

Proof, EXT2 P2 EQ-Bank 31

Using mathematical induction and integration by parts, show

  `int_0^pi sin^nx\ dx=(n-1)/n int_0^pi sin^(n-2)x\ dx`  for  `n>=2.`   (4 marks)

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  1. `text{Proof (See Worked Solution)`
Show Worked Solution

`text{Prove}\ \ int_0^pi sin^nx\ dx=(n-1)/n int_0^pi sin^(n-2)x\ dx\ \ text{for}\ \ n>=2.`

`text{If}\ \ n=2:`

`text{LHS}` `=int_0^pi sin^(2)x\ dx`  
  `=1/2 int_0^pi (1-cos (2x))\ dx`  
  `=1/2 [x-1/2sin(2x)]_0^pi`  
  `=1/2[(pi-1/2sin(2pi))-0]`  
  `=pi/2`  

 

`text{RHS}` `=(2-1)/2 int_0^pi sin^0 x\ dx`  
  `=1/2int_0^pi 1\ dx`  
  `=1/2[x]_0^pi`  
  `=pi/2`  
  `=\ text{LHS}`  

 
`:.\ text{True for}\ \ n=2`
 

`text{Assume true for}\ \ n=k`

`text{i.e.}\ int_0^pi sin^kx\ dx=(k-1)/k int_0^pi sin^(k-2)x\ dx`
 

`text{Prove true for}\ \ n=k+1`

`text{i.e.}\ int_0^pi sin^(k+1)x\ dx=(k)/(k+1) int_0^pi sin^(k-1)x\ dx`

`text{Let}\ \ I=\ text{LHS} = int_0^pi sin^(k+1)x\ dx=int_0^pi sinx * sin^(k)x\ dx`

`text{Using IBP:}`

`u` `=sin^k x` `v^(′)=sinx`
`u^(′)` `=kcosx*sin^(k-1)x`     `v=-cosx`

 

`I` `=[uv]_0^pi-int_0^pi u^(′)v\ dx`  
  `=[-sin^kx*cosx]_0^pi-int_0^pi -kcosx*sin^(k-1)x*cosx\ dx`  
  `=[(-sin^k pi*cos pi)-(-sin^k 0*cos 0)+kint cos^2x*sin^(k-1)x\ dx`  
  `=0+kint (1-sin^2x)*sin^(k-1)x\ dx`  
  `=kint_0^pi sin^(k-1)x\ dx-kint_0^pi sin^(k+1)x\ dx`  
`I` `=kint_0^pi sin^(k-1)x\ dx-kI`  
`I+kI` `=kint_0^pi sin^(k-1)x\ dx`  
`I(k+1)` `=kint_0^pi sin^(k-1)x\ dx`  
`I` `=k/(k+1) int_0^pi sin^(k-1)x\ dx`  
  `=\ text{RHS}`  

 
`:.\ text{True for}\ \ n=k+1`

`:.\ text{Since true for} \ n=2,\ text{by PMI, true for integers} \ n>=2.`

Proof, EXT2 P2 EQ-Bank 15

Using mathematical induction, prove that the maximum number of diagonals of an `n`-sided plane convex polynomial is `(n(n-3))/2` for `n>=4`.  (3 marks)

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`text{See Worked Solutions}`

Show Worked Solution

`text{Let}\ \ D_n=\ text{maximum number of diagonals of}\ n text{-sided polygon}`

`text{Prove true for}\ \ n=4:`

`D_4= 2\ \ text{(quadrilateral only has 2 possible diagonals)}`

`(4(4-3))/2 = 2`

`:.\ text{True for}\ \ n=4.`
 

`text{Assume true for}\ \ n=k:`

`text{i.e.}\ D_k=(k(k-3))/2`
 

`text{Prove true for}\ \ n=k+1:`

`text{i.e.}\ D_(k+1)=((k+1)(k+1-3))/2=((k+1)(k-2))/2`
 

`text{Let}\ ktext{-sided polygon be defined by points}\ P_1,P_2,…,P_k`

`text{Let}\ (k+1)text{-sided polygon be defined by points}\ P_1,P_2,…,P_k,P_(k+1)`

`=>\ text{The extra point adds}\ (k-1)\ text{diagonals}`
 

`D_(k+1)` `=D_k+(k-1)`  
  `=(k(k-3))/2+ (k-1)`  
  `=(k^2-3k+2k-2)/2`  
  `=(k^2-k-2)/2`  
  `=((k+1)(k-2))/2`  
  `=\ text{RHS}`  

 
`:.\ text{True for}\ \ n=k+1`

`:.\ text{Since true for}\ n=4,\ text{by PMI, true for integers}\ n>=4.`

Proof, EXT2 P2 EQ-Bank 14

Using mathematical induction, prove that the sum of the internal angles of an `n`-sided polynomial is  `180(n-2)°`  for  `n>=3`.  (3 marks)

Show Answers Only

`text{See Worked Solutions}`

Show Worked Solution

`text{Let}\ \ S_n=\ text{sum of interior angles of}\ n text{-sided polygon}`

`text{Prove true for}\ \ n=3:`

`S_3= 180°\ \ text{(sum of internal angles of a Δ)}`

`180(3-2) = 180°`

`:.\ text{True for}\ \ n=3.`
 

`text{Assume true for}\ \ n=k:`

`text{i.e.}\ S_k=(k-2)180°`
 

`text{Prove true for}\ \ n=k+1:`

`text{i.e.}\ S_(k+1)=(k+1-2)180°=(k-1)180°`
 

`text{Let}\ ktext{-sided polygon be defined by points}\ P_1,P_2,…,P_k`

`text{Let}\ (k+1)text{-sided polygon be defined by points}\ P_1,P_2,…,P_k,P_(k+1)`

`S_(k+1)` `=S_k+\ text{angle sum}\ ΔP_1P_kP_(k+1)`  
  `=(k-2)180° + 180°`  
  `=(k-1)180°`  
  `=\ text{RHS}`  

 
`:.\ text{True for}\ \ n=k+1`

`:.\ text{Since true for}\ n=3,\ text{by PMI, true for integers}\ n>=3.`

Proof, EXT2 P2 EQ-Bank 24

`n` lines are drawn in a 2-dimensional plane such that no three lines are concurrent and no two lines are parallel.

`S_n` is the number of regions into which these lines divide the plane with the diagram illustrating that `S_3=7`
 

  1. In a similar way, draw diagrams that illustrate `S_1, S_2` and `S_4`.  (1 mark)

  2. Using the results in part (i), make a conjecture about the general formula for `S_n`.  (2 marks)

  3. Use mathematical induction to prove the formula from part (ii).  (3 marks)

Show Answers Only

a.    `text{See Worked Solutions}`

b.    `S_n=(n^2+n+2)/2`

c.    `text{See Worked Solutions}`

Show Worked Solution

a.       `S_1=2`                        `S_2=4`
 
             
 

   `S_4=11`


 

b.    `text{Consider the pattern above:}`

`S_1=2,\ \ S_2=S_1+2`

`S_3=S_2+3,\ …\ , S_n=S_(n-1)+n`

`S_n` `=S_(n-1)+n`  
  `=S_(n-2)+(n-1)+n`  
  `=S_(n-3)+(n-2)+(n-1)+n`  
  `vdots`  
  `=S_1+(2+3+…+(n-1)+n)`  
  `=S_0+(1+2+…+(n-1)+n)`  
  `=1+(1+2+…+(n-1)+n)`  

 
`=>text{AP where}\ \ a=1, l=n, n=n`

`S_n` `=1+n/2(1+n)`  
  `=(2+n(n+1))/2`  
  `=(n^2+n+2)/2`  

 

c.    `text{Prove}\ \ S_n=(n^2+n+2)/2\ \ text{for}\ \ n>=0`

`text{If}\ \ n=1`

`text{LHS}\ =S_1=2`

`text{RHS}\ = (1^2+1+2)/2=2=\ text{LHS}`

`:.\ text{True for}\ n=1.`
 

`text{Assume true for}\ \ n=k:`

`text{i.e.}\ S_k=(k^2+k+2)/2`

`text{Prove true for}\ \ n=k+1:`

`text{i.e.}\ S_(k+1)=((k+1)^2+(k+1)+2)/2=(k^2+3k+4)/2`
 

`text{Consider the line}\ d_4\ text{added below that crosses 3 existing lines}`

`text{and creates 4 new regions.}`

`text{Similarly, the line}\ d_(k+1)\ text{will cross}\ k\ text{existing lines and}`

`text{create}\ (k+1)\ text{new regions.}`

`S_(k+1)` `=S_k+k+1`  
  `=(k^2+k+2)/2+k+1`  
  `=(k^2+k+2+2k+2)/2`  
  `=(k^2+3k+4)/2`  
  `=\ text{RHS}`  

 
`:.\ text{True for}\ \ n=k+1`

`:.\ text{Since true for}\ n=1,\ text{by PMI, true for integers}\ n>=1`

Proof, EXT2 P2 EQ-Bank 30

Using mathematical induction, prove that the sum of the external angles of an `n`-sided plane convex polygon equals 360° for  `n>=3`.   (4 marks)

Show Answers Only

`text{See Worked Solutions}`

Show Worked Solution

`text{Prove true for}\ \ n=3:`

 

`text{Let}\ \ S_3=\ text{sum of exterior angles for a 3-sided polynomial (triangle)}`

`S_3=angleACF+angleCBE+angleBAD`

`angleACF+angleACB=180°\ \ (angleFCB\ text{is a straight angle})`

`angleCBE+angleCBA=180°\ \ (angleABE\ text{is a straight angle})`

`angleBAD+angleBAC=180°\ \ (angleDAC\ text{is a straight angle})`

`angleACF+angleACB+angleCBE+angleCBA+angleBAD+angleBAC=540°`

 
`text{Internal angle sum of}\ ΔABC=180°:`

`angleACF+angleCBE+angleBAD+180°` `=540°`  
`angleACF+angleCBE+angleBAD` `=360°`  

 
`:.\ text{True for}\ \ n=3`
 

`text{Assume true for}\ \ n=k`

`text{i.e.}\ \ S_k=360°`

`text{Prove true for}\ \ n=k+1:`

`text{i.e.}\ \ S_(k+1)=360°`

`text{Let}\ ktext{-sided polygon be defined by points}\ P_1,P_2,…,P_k`

`text{Let}\ (k+1)text{-sided polygon be defined by points}\ P_1,P_2,…,P_k,P_(k+1)`
 

`S_k` `=theta_1+theta_2+ … +theta_(k-1)+theta_k`  
  `=theta_1+theta_2+ … +theta_(k-1)+alpha+beta`  

 
`S_(k+1)=mu+theta_2+ … +theta_(k-1)+beta+delta`

`theta_1=mu+phi\ \ text{(vertically opposite)}`

`=>\ \ mu=theta_1-phi\ …\ (1)`

`delta=alpha+phi\ …\ (2)\ \  text{(exterior angle of triangle)}`

  
`text{Substitute (1) and (2) into}\ S_(k+1):`

`S_(k+1)` `=theta_1-phi+theta_2+ … +theta_(k-1)+beta+alpha+phi`  
  `=theta_1+theta_2+ … +theta_(k-1)+alpha+beta`  
  `=S_k`  

 

`:.\ text{True for}\ \ n=k+1`

`:.\ text{Since true for}\ n=3,\ text{true for integers}\ n>=3.`

Calculus, EXT2 C1 2022 HSC 14b

Let  `J_(n)=int_(0)^(1)x^(n)e^(-x)\ dx`, where "n" is a non-negative integer.

  1. Show that  `J_(0)=1-(1)/(e)`.  (1  mark)

  2. Show that  `J_(n) <= (1)/(n+1)`.  (2 marks)

  3. Show that  `J_(n)=nJ_(n-1)-(1)/(e)`.  (2 marks)

  4. Using parts (i) and (iii), show by mathematical induction, or otherwise, that for all `n >= 1`,
  5.        `J_(n)=n!-(n!)/(e)sum_(r=0)^(n)(1)/(r!)`    (2 marks)

  6. Using parts (ii) and (iv) prove that  `e=lim_(n rarr oo)sum_(r=0)^(n)(1)/(r!)`.  (1  mark)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `text{Proof (See Worked Solutions)}`
  3. `text{Proof (See Worked Solutions)}`
  4. `text{Proof (See Worked Solutions)}`
  5. `text{Proof (See Worked Solutions)}`
Show Worked Solution
i.    `J_0` `=int_0^1 e^(-x)\ dx`
    `=[-e^(-x)]_0^1`
    `=-e^(-1)+1`
    `=1-1/e`

 


Mean mark (i) 93%.

ii.  `text{Show}\ \ J_n<=1/(n+1)`

`text{Note:}\ e^(-x)<1\ \ text{for}\ \ x in [0,1]`

`J_n` `=int_0^1 x^n e^(-x)\ dx`  
  `leq int_0^1 x^n \ dx`  
  `leq 1/(n+1)[x^(n+1)]_0^1`  
  `leq 1/(n+1)(1^(n+1)-0)`  
  `leq 1/(n+1)\ \ text{… as required}`  

 


♦♦ Mean mark (ii) 28%.
 

iii.  `text{Show}\ \ J_n=nJ_(n-1)-1/e`

`u` `=x^n` `v′` `=e^(-x)`
`u′` `=nx^(n-1)` `v` `=-e^(-x)`
`J_n` `=[-x^n * e^(-x)]_0^1-int_0^1 nx^(n-1)*-e^(-x)\ dx`  
  `=(-1^n * e^(-1)+0^n e^0)+nint_0^1 x^(n-1)*e^(-x)\ dx`  
  `=nJ_(n-1)-1/e`  

 
iv.   `text{Prove}\ \ J_(n)=n!-(n!)/(e)sum_(r=0)^(n)(1)/(r!)\ \ text{for}\ \ n >= 0`

`text{If}\ \ n=0:`

`text{LHS} = 1-1/e\ \ text{(see part (i))}`

`text{RHS} = 0!-0!/e (1/(0!)) = 1-1/e(1)=\ text{LHS}`

`:.\ text{True for}\ \ n=0.`
 

`text{Assume true for}\ \ n=k:`

`J_(k)=k!-(k!)/(e)sum_(r=0)^(k)(1)/(r!)`
   


♦ Mean mark (iv) 50%.

`text{Prove true for}\ \ n=k+1:`

`text{i.e.}\ \ J_(k+1)=(k+1)!-((k+1!))/(e)sum_(r=0)^(k+1)(1)/(r!)`

`J_(k+1)` `=(k+1)J_k-1/e\ \ text{(using part (iii))}`  
  `=(k+1)(k!-(k!)/(e)sum_(r=0)^(k)(1)/(r!))-1/e`  
  `=(k+1)!-((k+1)!)/(e)sum_(r=0)^(k)(1)/(r!)-1/e xx ((k+1)!)/((k+1)!)`  
  `=(k+1)!-((k+1)!)/e(\ sum_(\ r=0)^(k)(1)/(r!)+1/((k+1)!))`  
  `=(k+1)!-((k+1)!)/e(\ sum_(\ r=0)^(k+1)(1)/(r!))`  

 
`=>\ text{True for}\ \ n=k+1`

`:.\ text{S}text{ince true for}\ n=1,\ text{by PMI, true for integers}\ n>=1`
 

v.   `0<=J_n<= 1/(n+1)\ \ \ text{(part (ii))}`

`lim_(n->oo) 1/(n+1)=0\ \ => \ lim_(n->oo) J_n=0`

  
`text{Using part (iv):}`

`J_n/(n!)` `=1-1/e sum_(r=0)^(n)(1)/(r!)`  
`1/e sum_(r=0)^(n)(1)/(r!)` `=1-J_n/(n!)`  
`sum_(r=0)^(n)(1)/(r!)` `=e-(eJ_n)/(n!)`  
`lim_(n->oo)(\ sum_(\ r=0)^(n)(1)/(r!))`  `=lim_(n->oo)(e-(eJ_n)/(n!))`  
  `=e-0`  
  `=e`  

♦♦ Mean mark (v) 34%.

Proof, EXT2 P2 2022 HSC 13b

The numbers `a_(n)`, for integers `n >= 1`, are defined as
 

               `{:[a_(1)=sqrt2],[a_(2)=sqrt(2+sqrt2)],[a_(3)=sqrt(2+sqrt(2+sqrt2)) \ text{, and so on.}]:}`
 

These numbers satisfy the relation  `a_(n+1)^(2)=2+a_(n)`, for  `n >= 1`.     (Do NOT prove this)

Use mathematical induction to prove that  `a_(n)=2cos\ pi/(2^(n+1))`, for all integers  `n >= 1`.  (4 marks)

Show Answers Only

`text{Proof (See Worked Solution)}`

Show Worked Solution

`text{Prove}\ \ a_(n)=2cos(pi/(2^(n+1)))\ \ text{for}\ \ n >= 1`

`text{If}\ \ n=1:`

`a_1=2cos((pi)/(2^2))=2xx1/sqrt2=sqrt2`

`:.\ text{True for}\ n=1.`
 


Mean mark 58%.

`text{Assume true for}\ \ n=k:`

`a_(k)= =2cos(pi/(2^(k+1)))`
 

`text{Prove true for}\ \ n=k+1:`

`text{i.e.}\ \ a_(k+1)= 2cos(pi/(2^(k+2)))`

`text{LHS}` `=sqrt(2+a_k)`  
  `=sqrt(2+2cos(pi/(2^(k+1)))`  
  `=sqrt(2+2 cos(2 xx (pi/(2^(k+2)))))\ \ \ text{(using}\ \ cos(2theta)=2cos^2theta-1text{)}`  
  `=sqrt(2+2(2cos^2(pi/(2^(k+2)))-1)`  
  `=sqrt(2+4cos^2(pi/(2^(k+2)))-2)`  
  `=sqrt(4cos^2(pi/(2^(k+2)))`  
  `=2cos(pi/(2^(k+2)))`  
  `=\ text{RHS}`  

 

Proof, EXT2 P2 2021 HSC 12d

Prove by mathematical induction that  `sqrt{n!} > 2^n` , for integers  `n ≥ 9`.  (3 marks)

Show Answers Only

`text{See Worked Solutions}`

Show Worked Solution

`text{Prove} \ sqrt{n!} > 2^n \ text{for integers} \ n≥ 9`

`text{If} \ n = 9,`

`text(LHS) = sqrt{9!} = 602.39 …`

`text(RHS) = 2^9 = 512 < 602.39 …`

`:. \ text{True for} \ n = 1`
 

`text{Assume true for} \ n = k`

`text{i.e.} \ \ sqrt{k!} > 2^k`

`text{Prove true for} \ n = k +1`

`text{i.e.} \ \ sqrt{(k + 1)!} > 2^{k + 1}`
 

`text(LHS)` `= sqrt{(k + 1)!}`
  `= sqrt{k!} sqrt{k + 1}`
  `> 2^k * sqrt{k + 1}`
  `> 2^k * 2 \ \ \ (k ≥ 9\ => \ sqrt{k + 1} > 2)`
  `> 2^{k + 1}`
 
 
`=> \ text{True for} \ \ n = k + 1`
 
`:. \ text{S} text{ince true for} \ n = 9 , text{by PMI, true for integral} \ n ≥ 9.`

Proof, EXT2 P2 2020 HSC 14c

Prove by mathematical induction that, for  `n ≥ 2`,

`frac{1}{2^2} + frac{1}{3^2} + ... + frac{1}{n^2} < frac{n - 1}{n}`    (4 marks)

Show Answers Only

`text{See Worked Solutions}`

Show Worked Solution

`text{Prove} \ frac{1}{2^2} + frac{1}{3^2} + … + frac{1}{n^2} < frac{n – 1}{n}\ \ \ text(for)\ \ n>=2`
 

`text(If)\ \ n=2:`

`text{LHS} = frac{1}{2^2} = frac{1}{4}`

`text{RHS} = frac{2-1}{2} = frac{1}{2} > text{LHS}`
 
`therefore \ text{True for} \ \ n = 2`
 

`text{Assume true for} \ \ n =k`

`frac{1}{2^2} + frac{1}{3^2} + … + frac{1}{k^2} < frac{k – 1}{k}`
 

`text{Prove true for} \ \ n = k + 1`

`frac{1}{2^2} + frac{1}{3^2} + … + frac{1}{k^2} + frac{1}{(k + 1)^2}< frac{k}{k+1}`
 

`text{LHS}` `= frac{k – 1}{k} + frac{1}{(k + 1}^2}`
  `= frac{(k – 1)(k + 1)^2 + k}{k (k + 1)^2}`
  `= frac{(k – 1)(k^2 + 2 k +1) + k}{k(k +1)^2}`
  `= frac{k^3 + 2k^2 + k – k^2 – 2k – 1 + k}{k(k + 1)^2}`
  `= frac{k^3 + k^2 -1}{k(k + 1)^2}`
  `= frac{k^2 (k + 1 – frac{1}{k^2})}{k(k + 1)^2}`
  `= frac{k}{k +1} xx frac{(k + 1 – frac{1}{k^2})}{k +1}`
  `< frac{k}{k +1} \ \ \ (text{since} \ frac{k + 1 – frac{1}{k^2}}{k +1} < 1 )`

 
`=> \ text{True for} \ n = k + 1`

`therefore \ text{S}text{ince true for} \ n = 2, \ text{by PMI, true integral} \ n ≥ 2.`

Proof, EXT2 P2 2019 HSC 14c

  1. Show that  `cot x - cot 2x = text(cosec)\ 2x`.  (2 marks)

  2. Use mathematical induction to prove that, for all  `n >= 1`,

`sum_(r = 1)^n\ text(cosec)(2^r x) = cot x - cot(2^n x)`.  (2 marks)

Show Answers Only
  1. `text(Proof)\ text{(See Worked Solutions)}`
  2. `text(Proof)\ text{(See Worked Solutions)}`
Show Worked Solution

i.    `text(Show)\ \ cot x – cot 2x = text(cosec)\ 2x`

`text(LHS)` `= (cos x)/(sin x) – 1/(tan 2x)`
  `= (cos x)/(sin x) – (1 – tan^2 x)/(2 tan x)`
  `= (cos x)/(sin x) – ((1 – (sin^2 x)/(cos^2 x))/(2 (sin x)/(cos x)))`
  `= (cos x)/(sin x) – ((cos^2 x – sin^2 x)/(2 sin x cos x))`
  `= (2 cos^2 x – cos^2 x + sin^2 x)/(2 sin x cos x)`
  `= 1/(sin 2x)`
  `= text(cosec)\ 2x`
  `= ­text(RHS)`

 

ii.    `text(Prove)\ \ sum_(r = 1)^n\ text(cosec)(2^rx) = cot x – cot 2^n x\ \ text(for)\ \ n >= 1`

`text(Show true for)\ \ n = 1:`

♦ Mean mark 45%.

`text(LHS) = text(cosec)(2x)`

`text(RHS) = cot x – cot 2x = text(cosec)(2x)\ \ text{(using part (i))}`

`:.\ text(True for)\ \ n = 1`
 

`text(Assume true for)\ \ n = k:`

`text(cosec)\ 2x + text(cosec)\ 4x + … + text(cosec)\ 2^rx = cot x – cot 2^r x`

`text(Prove true for)\ \ n = k + 1:`

`text(i.e. cosec)\ 2x + … + text(cosec)\ 2^r x + text(cosec)\ 2^(r + 1) x = cot x – cot 2^(r + 1) x`

`text(LHS)` `= cot x – cot 2^r x + text(cosec)\ 2^(r + 1) x`
  `= cot x – cot 2^r x + text(cosec)\ (2.2^r x)`
  `= cot x – cot 2^r x + cot 2^r x – cot 2^(r + 1) x`
  `= cot x – cot 2^(r + 1) x`
  `= ­text(RHS)`

 
`:.\ text(True for)\ \ n=k+1`

`:.\ text(S)text(ince true for)\ \ n=1, text(by PMI, true for integral)\ \ n>=1.`

Proof, EXT2 P2 EQ-Bank 35

Use mathematical induction to prove that

`n^5 + n^3 + 2n`

is divisible by 4 for integers  `n >= 1.`   (4 marks)

Show Answers Only

`text(See Worked Solutions)`

Show Worked Solution

`text(If)\ n = 1,`

`1 + 1 + 2 xx 1 = 4\ \ text{(divisible by 4)}`

`:. text(True for)\ n = 1`
 

`text(Assume true for)\ n = k`

`text(i.e.)\ \ k^5 + k^3 + 2k = 4P\ \ …\ text{(1)}\ \ (text(where)\ P ∈\ text(integer))`
 

`text(Prove true for)\ n = k + 1`

`text(i.e.)\ \ (k + 1)^5 + (k + 1)^3 + 2(k + 1)\ \ text(is divisible by 4)`

`text(Expanding:)`

`k^5 + 5k^4 + 10k^3 + 10k^2 + 5k + 1 + k^3 + 3k^2 + 3k + 1 + 2k + 2`

`= k^5 + 5k^4 + 11k^3 + 13k^2 + 8k + 4`

`= 4P + 5k^4 + 10k^3 + 13k^2 + 8k + 4\ \ \ \ text{(see (1) above)}`

`= 4P + 4 underbrace{(k^4 + 2k^3 + 3k^2+2k + 1)}_(text(integer)\ Q) + k^4 + 2k^3 + k^2`

`= 4(P + Q) + k^2(k^2 + 2k + 1)`

`= 4(P + Q) + k^2(k + 1)^2`
 

`text(For any integer)\ \ k >= 2, \ k^2(k + 1)^2\ \ text{is  (odd)}^2 xx text{(even)}^2`

`text{and  (even)}^2 = (2R)^2\ \ text(where)\ \ R ∈\ text(integer)`

`=> k^2(k + 1)^2 = 4R^2 xx (text(odd))^2\ \ \ text{(divisible by 4)}`

`:. text(True for)\ \ n = k + 1`

`:. text(S)text(ince true for)\ \ n = 1,\ text(by PMI, true for integral)\ n >= 1.`

Proof, EXT2 P2 2018 HSC 16a

Use mathematical induction to prove that, for  `n >= 1`,

`x^((3^n)) - 1 = (x - 1)(x^2 + x + 1)(x^6 + x^3 + 1) … (x^((2 xx 3^(n - 1))) + x^((3^(n - 1))) + 1)`.  (3 marks)

Show Answers Only

`text(See Worked Solutions)`

Show Worked Solution

`text(If)\ \ n = 1`

♦ Mean mark 51%.

`text(LHS) = x^3 – 1`

`text(RHS)` `=(x-1)(x^2+x+1)`  
  `=x^3+x^2+x-x^2-x-1`  
  `=x^3-1 = text(LHS)`  

 
`:.\ text(True for)\ n = 1`
 

`text(Assume true for)\ \ n = k`

`x^(3^k) – 1 = (x – 1)(x^2 + x + 1) … (x^(2 xx 3^(k – 1)) + x^(3^(k – 1)) + 1)`

 
`text(Prove true for)\ \ n = k + 1`

`text(i.e.)\ \ \ x^(3^(k + 1)) = underbrace{(x – 1)(x^2 + x + 1) …(x^(2 xx 3^(k – 1)) + x^(3^(k – 1)) + 1)}_{x^(3^k) – 1) (x^(2 xx 3^k) + x^(3^k) + 1)`

`x^(3^(k + 1)) – 1` `= (x^(3^k) – 1)(x^(2 xx 3^k) + x^(3^k) + 1)`
  `= x^(3 xx 3^k) + x^(2 xx 3^k) + x^(3^k) – x^(2 xx 3^k) – x^(3^k) – 1`
  `= x^(3^(k + 1)) – 1`

 
`=>\ text(True for)\ \ n = k + 1`

`:.\ text(S)text(ince true for)\ \ n = 1,\ text(by PMI, true for integral)\ \ n >= 1.`

Proof, EXT2 P2 2017 HSC 16c

A 2 by `n` grid is made up of two rows of `n` square tiles, as shown.
 

The tiles of the 2 by `n` grid are to be painted so that tiles sharing an edge are painted using different colours. There are `x` different colours available, where  `x ≥ 2`.

It is NOT necessary to use all the colours.

Consider the case of the 2 by 2 grid with tiles labelled A, B, C and D, as shown.
 

There are `x(x - 1)` ways to choose colours for the first column containing tiles A and B. Do NOT prove this.

  1. Assume the colours for tiles A and B have been chosen. There are two cases to consider when choosing colours for the second column. Either tile C is the same colour as tile B, or tile C is a different colour from tile B.

     

    By considering these two cases, show that the number of ways of choosing colours for the second column is  `x^2 - 3x +3`.  (2 marks)

  2. Prove by mathematical induction that the number of ways in which the 2 by `n` grid can be painted is  `x(x - 1)(x^2 - 3x + 3)^(n - 1)`, for  `n ≥ 1`.  (2 marks)

  3. In how many ways can a 2 by 5 grid be painted if 3 colours are available and each colour must now be used at least once?  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `480`
Show Worked Solution
i.   

`text(If Colour)\ C =\ text(Colour)\ B:`

♦♦♦ Mean mark 20%.

`text(Column 2 combinations)`

`= 1 xx (x – 1)`

`= x – 1`

 

`text(If Colour)\ C !=\ text(Colour)\ B:`

`text(Column 2 combinations)`

`= (x – 2)(x – 2)`

 

`:.\ text(Total number of ways for column 2)`

`= x – 1 + (x – 2)^2`

`= x^2 – 3x + 3\ \ …\ text(as required.)`

 

ii.   `P(n) = x(x – 1)(x^2 – 3x + 3)^(n – 1)`

♦♦♦ Mean mark 22%.

`text(If)\ \ n = 1\ \ (text(i.e. 1 column)),`

`text(Combinations) = x(x – 1)\ \ (text(given))`

`P(1)` `= x(x – 1)(x^2 – 3x + 3)^0`
  `= x(x – 1)`

`:. text(True for)\ n = 1`

 

`text(Assume true for)\ n = k:`

`text(i.e. Possible combinations for)\ k\ text(columns)`

`P(k) = x(x – 1)(x^2 – 3x + 3)^(k – 1)`

`text(Prove true for)\ \ n = k + 1`

`text(i.e.)\ \ P(k + 1) = x(x – 1)(x^2 – 3x + 3)^k`

 

`text(Consider a grid)\ 2 xx k\ text(and add an extra two)`

`text(tiles to make a)\ 2 xx (k + 1)\ text(grid.)`

`text(Total possible combinations)`

`= text(combinations of)\ (2 xx k)\ text(grid) xx (x^2 – 3x + 3)\ \ (text{from (i)})`

`= x(x – 1)(x^2 – 3x + 3)^(k – 1) xx (x^2 – 3x + 3)`

`= x(x – 1)(x^2 – 3x + 3)^k`

`=> text(True for)\ \ n = k + 1`

`:. text(S)text(ince true for)\ n = 1,\ text(by PMI, true for integral)\ n >= 1.`

 

iii.   `text(If)\ \ n = 5, \ x = 3:`

♦♦♦ Mean mark 22%.

`text{Total combinations (includes using only 2 colours)}`

`P(5)` `= 3(3 – 1)(3^2 – 3.3 + 3)^4`
  `= 3(2)(3)^4`
  `= 486`

 

`text(Consider pattern when only 2 colours used:)`

`text(Combinations that only use 2 colours)`

`=\ text(Colours in box 1 × possible colours in box 2)`

`= 3 xx 2`

`= 6`

`:.\ text(Combinations if each colour used at least once)`

`= 486 – 6`
`= 480`

Proof, EXT2 P2 2016 HSC 16c

In a group of `n` people, each has one hat, giving a total of `n` different hats. They place their hats on a table. Later, each person picks up a hat, not necessarily their own.

A situation in which none of the `n` people picks up their own hat is called a derangement.

Let  `D(n)`  be the number of possible derangements.

  1. Tom is one of the `n` people. In some derangements Tom finds that he and one other person have each other's hat.

     

    Show that, for  `n > 2`, the number of such derangements is  `(n - 1) D (n - 2).`  (1 mark)

  2. By also considering the remaining possible derangements, show that, for  `n > 2,`

    `qquad qquad D(n) = (n - 1) [D(n - 1) + D(n - 2)].`  (2 marks)

  3. Hence, show that  `D(n) - nD(n - 1) = -[D(n - 1) - (n - 1) D(n - 2)]`,  for  `n > 2.`  (1 mark)

  4. Given  `D(1) = 0`  and  `D(2) = 1`,  deduce that  `D(n) - n D(n - 1) = (-1)^n`,  for  `n > 1.`  (1 mark)

  5. Prove by mathematical induction, or otherwise, that for all integers  `n >= 1,\ D(n) = n! sum_(r = 0)^n (-1)^r/(r!).`  (2 marks)

Show Answers Only

a.    `text(See Worked Solutions)`

b.    `text(See Worked Solutions)`

c.    `text(See Worked Solutions)`

d.    `text(See Worked Solutions)`

e.    `text(See Worked Solutions)`

Show Worked Solution

a.    `text(Tom and one other person can have each)`

♦♦♦ Mean mark 22%.

`text(other’s hat in)\ (n – 1)\ text(combinations.)`

`text(There are)\ \ D(n – 2)\ \ text(possibilities for the)`

`text(rest of the people selecting the wrong hats.)`

`:. text(Number of derangements) = (n – 1)D(n – 2)`

 

b.    `text(Consider all possible derangements:)`

♦♦♦ Mean mark 6%.

`text(Any of)\ n\ text(people choose the wrong hat.)`

`text(Remainder)\ (n – 1)\ text(people can select the)`

`text(wrong hat in)\ D(n – 1)\ text(ways.)`

`:. nD(n – 1)\ text(derangements.)`

`text{This includes part (i) combinations.}`

 

`:.\ text(Remaining possible derangements)`

`= nD(n – 1) – (n – 1)D(n – 2)`

`= nD(n – 1) – D(n – 1)`

`= (n – 1)D(n – 1)`

`:. D(n)` `= (n – 1)D(n – 1) + (n – 1)D(n – 2)`
  `= (n – 1)[D(n – 1) + D(n – 2)]`

 

♦♦ Mean mark part (iii) 37%.
c.     `D(n)` `= (n – 1)[D(n – 1) + D(n – 2)]`
    `= nD(n – 1) – D(n – 1) + (n – 1)D(n – 2)`

 

`:. Dn – nD(n – 1) = −[D(n – 1) – (n – 1)D(n – 2)]`

 

d.    `D(1) = 0, D(2) = 1`

♦♦♦ Mean mark 1%!

`D(2) – 2D(1) = 1`

`D(3) – 3D(2) = −[D(2) – 2D(1)] = −1`

`D(4) – 4D(3) = −[D(3) – 3D(2)] = −(−1) = 1`

`D(5) – 5D(4) = −[D(4) – 4D(3)] = −1`

 

`:. D(n) – nD(n – 1) = (−1)^n\ text(for)\ n > 1`

 

e.    `text(Prove)\ D(n) = n! sum_(r = 0)^n ((−1)^r)/(r!)\ text(for)\ n >= 1`

`text(When)\ n = 1,`

`D(1) = 1! sum_(r = 0)^1 ((−1)^r)/(r!) = 1 – 1 = 0`

`text(S)text(ince)\ D(1) = 0\ (text(given)),`

`:.\ text(True for)\ n = 1`

♦♦♦ Mean mark 13%.

 

`text(Assume true for)\ \ n = k,`

`text(i.e.)\ \ D(k) = k! sum_(r = 0)^k ((−1)^r)/(r!)`

`text(Prove true for)\ \ n=k+1,`

`text(i.e.)\ \ D(k+1) = (k+1)!sum_(r = 0)^(k+1) ((−1)^r)/(r!)`

`D(k+1)`

`= (k + 1)D(k) + (−1)^(k + 1)qquad(text{from part (iv)})`
`= (k + 1) · k! sum_(r = 0)^k ((−1)^r)/(r!) + (−1)^(k + 1)`
`= (k + 1)!(1 – 1/(1!) + 1/(2!) – 1/(3!) + … + ((−1)^k)/(k!)) + (−1)^(k + 1) · ((k + 1)!)/((k + 1)!)`
`= (k + 1)!(1 – 1/(1!) + 1/(2!) – 1/(3!) + … + ((−1)^k)/(k!) + ((−1)^(k+1))/((k+1)!))`
`= (k + 1)! sum_(r = 0)^(k + 1) ((−1)^r)/(r!)`

 

`=> text(True for)\ n = k + 1.`

`:. text(S)text(ince true for)\ \ n = 1,\ text(by PMI, true for integral)\ n >= 1.`

Proof, EXT2 P2 2006 HSC 7c

The sequence  `{x_n}`  is given by

`x_1 = 1`  and  `x_(n + 1) = (4 + x_n)/(1 + x_n)`  for  `n >= 1.`

  1. Prove by induction that for  `n >= 1`,  `x_n = 2 ((1 + alpha^n)/(1 - alpha^n))`, where  `alpha = -1/3.`  (4 marks)

  2. Hence find the limiting value of  `x_n`  as  `n -> oo.`  (1 mark)

Show Answers Only

a.    `text(Proof)\ \ text{(See Worked Solutions)}`

b.    `2`

Show Worked Solution

a.    `text(If)\ \ n = 1`

`x_1=2 ((1 – 1/3)/(1 + 1/3))=(2 xx 2/3)/(4/3)=1`

`:.text(True for)\ \ n = 1`

 

`text(Assume true for)\ \ n=k,`

`text(i.e.)\ \ \ \ x_k = 2 ((1 + alpha^k)/(1 – alpha^k))`

`text(Prove true for)\ \ n=k+1`

`text(i.e.)\ \ \ \ x_(k + 1) = 2 ((1 + alpha^(k + 1))/(1 – alpha^(k+1)))`

`x_(k + 1) ­=` `(4 + x_k)/(1 + x_k)`
`­=` `(4 + 2 ((1 + alpha^k)/(1 – alpha^k)))/(1 + 2((1 + alpha^k)/(1 – alpha^k)))`
`­=` `2 [(2(1 – alpha^k) + (1 + alpha^k))/(1 – alpha^k + 2 (1 + alpha^k))]`
`­=` `2 [(2 – 2 alpha^k + 1 + alpha^k)/(1 – alpha^k + 2 + 2 alpha^k)]`
`­=` `2 [(3 – alpha^k)/(3 + alpha^k)]`
`­=` `2 [(1 – alpha^k xx 1/3)/(1 + alpha^k xx 1/3)]`
`­=` `2 [(1 + alpha^k xx alpha)/(1 – alpha^k xx alpha)],\ \ \ \ (alpha=-1/3)`
`­=` `2 [(1 + alpha^(k + 1))/(1 – alpha^(k + 1))]`
`­=` `text(RHS)`

 

`:.text(True for)\ \ n = k + 1\ \ text(if it is true for)\ \ n = k.`

`:.text(S)text(ince true for)\ \ n=1, text(by PMI, true for all integral)\ \ n >= 1.`

 

b.   `text(S)text(ince)\ \ lim_(n -> oo) (-1/3)^n=0`

`:.lim_(n -> oo) x_n` `=2 ((1 + (-1/3)^n)/(1 – (-1/3)^n))`
  `=2`

Proof, EXT2 P2 2007 HSC 6a

  1. Use the binomial theorem  `(a + b)^n = a^n + ((n), (1)) a^(n-1) b + … + b^n`
  2. to show that, for  `n >= 2,\ \ 2^n > ((n), (2)).`   (1 mark)
  1. Hence show that, for  `n >= 2`,
  2. `(n + 2)/2^(n-1) < (4n + 8)/(n(n-1)).`   (2 marks)

  3. Prove by induction that, for integers  `n >= 1`,
  4. `1 + 2 (1/2) + 3 (1/2)^2 + … + n (1/2)^(n-1) = 4-(n + 2)/2^(n-1).`   (3 marks)

  5. Hence determine the limiting sum of the series
  6. `1 + 2 (1/2) + 3 (1/2)^2 + ….`   (1 mark)

Show Answers Only

a.    `text(Proof)\ \ text{(See Worked Solutions)}`

b.    `text(Proof)\ \ text{(See Worked Solutions)}`

c.    `text(Proof)\ \ text{(See Worked Solutions)}`

d.    `4`

Show Worked Solution

a.    `text(Let)\ \ a = 1,\ \ b = 1`

`2^n` `= 1 + ((n), (1)) + ((n), (2)) + … + ((n), (n))`
`:. 2^n` `> ((n), (2))\ \ \ \ text(for)\ n>=2`

 

b.     `((n), (2))` `= (n(n-1))/(2 xx 1)`
  `:.2^n` `>(n(n-1))/(2 xx 1),\ \ \ \ \ text{(part (i))}`
  `1/2^n` `<2/(n (n-1))`
  `2/2^n` `<4/(n (n-1))`
  `1/2^(n-1) ` `<4/(n (n-1))`
  `:.(n + 2)/2^(n-1)` `< (4n + 8)/(n (n-1)),\ \ \ \ \ (n+2>0)`

 

c.    `text(If)\ n = 1:`

`text(LHS) = 1`

`text(RHS) = 4-(1+2)/2^0 = 1 = text(LHS)`

`:.\ text(True for)\ \ n = 1`
 

`text(Assume result is true for)\ \ n = k:`

`1 + 2 (1/2) + 3 (1/2)^2 + … + k (1/2)^(k-1) = 4-(k + 2)/2^(k-1).`
 

`text(Prove result is true for)\ \ n = k + 1:`

`1 + 2 (1/2) +  … + k (1/2)^(k-1) + (k + 1) (1/2)^k = 4-(k + 3)/2^k`

`text(LHS)` `=1 + 2 (1/2) + … + k (1/2)^(k-1) + (k + 1) (1/2)^k`
  `=4-(k + 2)/2^(k-1) + (k + 1)/2^k`
  `=4-(2k + 4-k-1)/2^k`
  `=4-(k + 3)/2^k`
  `=\ text(RHS)`

 

`=>text(True for)\ \ n = k + 1\ \ text(if it is true for)\ \ n = k.`

`:.\ text(S)text(ince true for)\ \ n = 1, text(by PMI, true for integers)\ \ n>=1`

 

d.    `lim_(n -> oo) (4-(n + 2)/2^(n-1))=4-lim_(n -> oo) ((n + 2)/2^(n-1))`

`text{Using part (b):}`

`(n + 2)/2^(n-1)` `< (4n + 8)/(n (n-1))`
`lim_(n -> oo) ((4n + 8)/(n (n-1)))` `= 0`
`:. lim_(n -> oo) ((n + 2)/2^(n-1)) ` `= 0`

 
`:.lim_(n -> oo) (4-(n + 2)/2^(n-1)) = 4-0 = 4`

Proof, EXT2 P2 2009 HSC 8a

  1. Using the substitution  `t = tan\­ theta/2`, or otherwise, show that
     
    `qquad cot theta + 1/2 tan\­ theta/2 = 1/2 cot\­ theta/2.`  (2 marks)

  2. Use mathematical induction to prove that, for integers  `n >= 1`,
     
    `qquad sum_(r = 1)^n 1/2^(r - 1) tan­ x/2^r = 1/2^(n - 1) cot\­ x/2^n - 2 cot x.`  (3 marks) 

  3. Show that  `lim_(n -> oo) sum_(r = 1)^n 1/2^(r - 1) tan\­ x/2^r = 2/x - 2 cot x.`  (2 marks)

  4. Hence find the exact value of

  5.  

    `qquad tan\ pi/4 + 1/2 tan\ pi/8 + 1/4 tan\ pi/16 + ….`  (2 marks)

Show Answers Only

a.    `text(Proof)\ \ text{(See Worked Solutions)}`

b.    `text(Proof)\ \ text{(See Worked Solutions)}`

c.    `text(Proof)\ \ text{(See Worked Solutions)}`

d.    `4/pi`

Show Worked Solution

a.    `t=tan\ theta/2, \ \ \ sin theta=(2t)/(1+t^2),\ \ \ cos theta=(1-t^2)/(1+t^2)`

`text(Prove)\ \ cot theta + 1/2 tan\­ theta/2 = 1/2 cot\­ theta/2`

`text(LHS)` `=cot theta + 1/2 tan\ theta/2`
  `=cos theta/sin theta+ 1/2 tan\ theta/2`
  `=(1-t^2)/(2t)+t/2`
  `=(1-t^2+t^2)/(2t)`
  `=1/(2t)`
  `=1/2 cot\ theta/2`
  `=\ text(RHS)`

 

b.    `text(If)\ \ n = 1`

`text(LHS)` `=1/2^0 tan­\ x/2^1=tan\ x/2`
`text(RHS)` `=1/2^0 cot­\ x/2 – 2 cot x`
  `=cot\ x/2 – 2 cot x`
`text{Using part (i)},\ \ 1/2 tan\ theta/2` `= 1/2 cot\ theta/2 – cot theta,`
`:.tan\ theta/2` `= cot­\ theta/2 – 2 cot theta`
`text(RHS)` `=tan\ x/2`
  `=\ text(LHS)`
`:.\ text(True for)\ \ n=1`

 

`text(Assume true for)\ \ n = k`

`text(i.e.)\ \ sum_(r = 1)^k 1/2^(r – 1) tan­\ x/2^r = 1/2^(k – 1) cot­\ x/2^k – 2 cot x.`

`text(Prove the result true for)\ \ n = k+1`

`text(i.e.)\ \sum_(r = 1)^(k + 1) 1/2^(r – 1) tan­ x/2^r = 1/2^k cot­ x/2^(k + 1) – 2 cot x`

`text(LHS)` `=sum_(r = 1)^(k) 1/2^(r – 1) tan­­ x/2^r + 1/2^k tan­ x/2^(k + 1)`
  `=1/2^(k – 1) cot­ x/2^k – 2 cot x + 1/2^k tan­ x/2^(k + 1)`
  `=1/2^(k – 1) (cot­ x/2^k + 1/2 tan­ x/2^(k +1)) – 2 cot x,\ \ \ \ text{(Let}\ \ theta=x/2^ktext{)}`
  `=1/2^(k – 1)(cot­\ theta + 1/2 tan\ theta/2) – 2 cot x`
  `=1/2^(k – 1)(1/2 cot\ theta/2) – 2 cot x,\ \ \ \ text{(from part (i))}`
  `=1/2^k cot\ theta/2 – 2 cot x`
  `=1/2^k cot­ x/2^(k + 1) – 2 cot x`
  `=\ text(RHS)`

 

`=>text(True for)\ \ n = k + 1\ \ text(if it is true for)\ \ n = k.`

`:.text(S)text(ince true for)\ \ n = 1,\ text(by PMI, true for integral)\ \ n >= 1.`

 

c.    `lim_(n -> oo) sum_(r = 1)^n 1/2^(r – 1) tan­ x/2^r `

`=lim_(n -> oo) (1/2^(n-1) cot­ x/2^n – 2 cot x)`

`=lim_(n -> oo) (2/x * x/2^n * 1/(tan­ x/2^n) – 2 cot x)`

`=2/x xx lim_(n -> oo) ((x/2^n)/(tan­ x/2^n)) – 2 cot x`

 

  `=>text(As)\ \ n -> oo,\ \ x/2^n=theta->0, and`

  `=>lim_(theta-> 0) (theta)/(tan­ theta) =1`

`:.lim_(n -> oo) sum_(r = 1)^n 1/2^(r – 1) tan­ x/2^r =2/x – 2 cot x`

 

d.    `tan\ pi/4 + 1/2 tan\ pi/8 + 1/4 tan\ pi/16 + …`

`=lim_(n -> oo) sum_(r = 1)^n 1/2^(r – 1) tan­ (pi/2)/2^r`

`=(2/(pi/2)) – 2 cot­ pi/2`

`=4/pi`

Proof, EXT2 P2 2010 HSC 6b

A sequence `a_n` is defined by  

`a_n = 2a_(n − 1) + a_(n − 2)`,

for `n ≥ 2`, with `a_0 = a_1 = 2`.

Use mathematical induction to prove that

`a_n = (1 + sqrt2)^n + (1 − sqrt2)^n`  for all  `n ≥ 0`.   (3 marks)

Show Answers Only

`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text(Prove)\ \ a_n = (1 + sqrt2)^n + (1 − sqrt2)^n\ text(for all)\ n ≥ 0.`

`text(where)\ a_0 = a_1 = 2, and  \ a_n = 2a_(n − 1) + a_(n − 2)\ \ text(for)\ \ n ≥ 2,\ `

 

`text(When)\ \ n=0,` `a_0` `= (1 + sqrt2)^0 + (1 − sqrt2)^0`
    `= 1 + 1=2`
`text(When)\ \ n=1,` `a_1` `= (1 + sqrt2)^1 + (1 − sqrt2)^1`
    `= 1 + sqrt2 + 1 − sqrt2=2`

 
`:.\ text(True for)\ n=0\ \ and\ n=1`

♦ Mean mark 36%.

 

`text(Assume that)`

`a_k = (1 + sqrt2)^k + (1 − sqrt2)^k\ text(and)`

`a_(k − 1) = (1 + sqrt2)^(k − 1) + (1 − sqrt2)^(k − 1)\ text(for all)\ k ≥ 1.`

 

`text(Prove true for)\ \ n = k + 1`

`text(i.e.)\ \ a_(k + 1) = (1 + sqrt2)^(k + 1) + (1 − sqrt2)^(k + 1).`

`a_(k + 1)` `= 2a_k + a_(k − 1).`
  `= 2(1 + sqrt2)^k + 2(1 − sqrt2)^k + (1 + sqrt2)^(k − 1) + (1 − sqrt2)^(k − 1)`
  `= (1 + sqrt2)^(k − 1)(2 + 2sqrt2 + 1) + (1 − sqrt2)^(k − 1)(2 − 2sqrt2 + 1)`
  `= (1 + sqrt2)^(k − 1)(1 + sqrt2)^2 + (1 − sqrt2)^(k − 1)(1− sqrt2)^2`
  `= (1 + sqrt2)^(k + 1) + (1 − sqrt2)^(k + 1)`

 

`=>\ text(True for)\ \ n = k + 1\ \ text(if it is true for)\ \ n = k\ \ text(and)\ \ n = k − 1.`

`:.\ text(S)text(ince true for)\ n = 0 and n=1,\ text(by PMI, it is)`

`text(true for integral)\ n ≥ 0.`

Proof, EXT2 P2 2011 HSC 3c

Use mathematical induction to prove that  `(2n)! >= 2^n (n!)^2`  for all positive integers  `n`.  (3 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

`text(If)\ \ n=1,`

`text(LHS) = 2! = 2`

`text(RHS) = 2 xx (1!)^2 = 2 = text(LHS)`
 

`text(Assume true for)\ \ n = k,`

  `text(i.e.)\ \ (2k)! >= 2^k (k!)^2\ \ \ …\ text{(i)}`
 

`text(Prove true for)\ \ n = k +1,`

  `text(i.e.)\ \ (2(k +1))! >= 2^(k + 1) ((k +1)!)^2`

`text(LHS)` `= (2 (k + 1))!`
  `= (2k + 2)(2k +1) xx underbrace{(2k)!}_text{using part (i)}`
  `>= (2k + 2) (2k +1) xx 2^k (k!)^2`
  `>= 2(k + 1) (2k + 1) xx 2^k (k!)^2`
  `>=2^(k+1) (k+1)(k+1) (k!)^2,\ \ \ \ \ (2k+1)>(k+1),\ text(for)\ k>0`
  `>= 2^(k + 1) ((k + 1)!)^2`

 

`=>\ text(True for)\ \ n = k + 1`

`:.\ text(S)text(ince true for)\ \ n = 1,\ text(by PMI, true for integral)\ n>=1.`

Proof, EXT2 P2 2012 HSC 16b

  1. Show that  `tan^(-1)\ x + tan^(-1)\ y = tan^(-1)((x + y)/(1 − xy))`  for  `|\ x\ | < 1`  and  `|\ y\ | < 1`.  (1 mark)

  2. Use mathematical induction to prove 
     
        `sum_(j = 1)^n\ tan^(-1)(1/(2j^2)) = tan^(-1)(n/(n + 1))`  for all positive integers `n`.  (3 marks)

  3. Find  `lim_(n → ∞) sum_(j = 1)^n\ tan^(-1)(1/(2j^2))`.  (1 mark)

Show Answers Only

a.    `text{Proof (See Worked Solutions.)}`

b.    `text{Proof (See Worked Solutions.)}`

c.    `pi/4`

Show Worked Solution

a.    `text(Let)\ \ A=tan^(-1)\ x,  and  B=tan^(-1)y`

`=> tan\ A=x,  and  tan\ B=y`

 

`tan\ (A+B)` `=(tan A + tan B)/(1-tanAtanB)`
  `=(x+y)/(1-xy)`
`:. A+B` `= tan^(-1)\ ((x+y)/(1-xy))\ \ \ text(… as required)`

 

b.    `sum_(j = 1)^n\ tan^(-1)\ (1/(2j^2)) = tan^(-1)\ (n/(n + 1))`

`text(If)\ \ j=1`

`text(LHS)` `=tan^(-1)\ (1/2)`
`text(RHS)` `=tan^(-1)\ (1/(1 + 1))`
  `=tan^(-1)\ (1/2)`
  `=\ text(LHS)`

 
`=>\ text(True for)\ \ j = 1.`

 

`text(Assume true)`

`sum_(j = 1)^n\ tan^(-1)\ (1/(2j^2)) = tan^(-1)\ (n/(n + 1))`

`text(Need to prove)`

`sum_(j = 1)^(n + 1)\ tan^(-1)\ (1/(2j^2)) = tan^(-1)\ ((n + 1)/(n + 2))`

`text(LHS)` `=sum_(j = 1)^(n + 1)\ tan^(-1)\ (1/(2j^2))`
  `=tan^(-1)\ (n/(n + 1)) + tan^(-1)\ (1/(2(n + 1)^2))`
  `=tan^(-1)\ ((n/(n + 1) + 1/(2(n + 1)^2))/(1 − n/(n + 1) xx 1/(2(n + 1)^2))),\ \ \ text{(using part (i))}`
  `=tan^(-1)\ ((2n(n + 1)^2 + n + 1)/(2(n + 1)^3 − n))`
  `=tan^(-1)\ (((n + 1)(2n^2 + 2n + 1))/(2(n + 1)^3 − n))`
  `=tan^(-1)\ (((n + 1)(2n^2 + 2n + 1))/(2n^3 + 6n^2 + 6n + 2 − n))`
  `=tan^(-1)\ (((n + 1)(2n^2 + 2n + 1))/(2n^3 + 6n^2 + 5n + 2))`
  `=tan^(-1)\ (((n + 1)(2n^2 + 2n + 1))/((n + 2)(2n^2 + 2n + 1)))`
  `=tan^(-1)\ ((n + 1)/(n + 2))`
  `=\ text(RHS)`

 

`=> text(True for)\ \ j=n+1`

`:.text(S)text(ince true for)\ \ j = 1,\ text(by PMI, true for integral)\ \ j>=1`

 

c.     `lim_(n → ∞) sum_(j = 1)^n\ tan^(-1)\ (1/(2j^2))`  `= lim_(n → ∞)\ tan^(-1)\ (n/(n + 1))`
    `= lim_(n → ∞)\ tan^(-1)\ (1/(1 + 1/n))`
    `= tan^(-1)\ 1`
    ` = pi/4`

Proof, EXT2 P2 2013 HSC 14b

Let  `z_2 = 1 + i`  and, for  `n > 2`, let  `z_n = z_(n - 1) (1 + i/(|\ z_(n - 1)\ |)).`
 

Use mathematical induction to prove that  `|\ z_n\ | = sqrt n`  for all integers  `n >= 2.`  (3 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution
♦ Mean mark 50%.

MARKER’S COMMENT: Many students did not recognise the need to use `|\ ab\ |=|\ a\ |*|\ b\ |` in their solution.

`text(Prove)\ \ |\ z_n\ | = sqrt n\ \ \ text(for)\ n>=2`

`text(given)\ \ z_n = z_(n – 1) (1 + i/(|\ z_(n – 1)\ |))`

`text(If)\ \ n=2`

` z_2 = 1 + i`

`|\ z_2\ | = sqrt (1 + 1) = sqrt 2`

`:.text(True for)\ \ n = 2`

 

`text(Assume that)\ \ |\ z_k\ | = sqrt k\ \ text(and prove that)\ \ |\ z_(k + 1)\ | = sqrt (k + 1)`

`z_k` `= z_(k – 1) (1 + i/(|\ z_(k – 1)\ |))`
`=> z_(k + 1) ` `= z_k (1 + i/(|\ z_k\ |))`

 

`|\ z_(k + 1)\ |` `= |\ z_k (1 + i/(|\ z_k\ |))\ |`
  `=|\ z_k\ |*|\ (1 + i/sqrt k)\ |`
  `= sqrt k * sqrt((1^2+(1/sqrt k)^2))`
  `=sqrt k * sqrt((1+1/k))`
  `= sqrt k * sqrt((k+1)/k)`
  `= sqrt k * sqrt(k+1)/sqrt k`
  `=sqrt(k+1)`

 

`=>text(True for)\ \ n = k + 1`

`:. text(S)text(ince true for)\ \ n=2,\ \ text(by PMI, true for all integers)\ \ n >= 2.`

Proof, EXT2* P2 2006 HSC 5d

  1. Use the fact that  `tan ( alpha-beta) = (tan alpha-tan beta)/(1 + tan alpha tan beta)`
    to show that

   `1 + tan\ n theta\ tan (n + 1) theta = cot theta (tan (n + 1) theta-tan\ n theta).`   (1 mark)

  1. Use mathematical induction to prove that, for all integers `n >= 1`

   `tan theta\ tan 2 theta + tan 2 theta\ tan 3 theta + … + tan\ n theta\ tan (n + 1) theta`

`=-(n + 1) + cot theta\ tan (n + 1) theta.`   (3 marks)

Show Answers Only

a.    `text(Proof)\ \ text{(See Worked Solutions)}`

b.    `text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

a.   `text(Show)`

`1 + tan\ n theta\ tan ( n + 1) theta = cot theta\ (tan (n+1) theta-tan\ n theta)`

`text(LHS)` `= underbrace{1 + tan\ n theta\ tan ( n + 1) theta}_text{rearrange part (i) identity}`
  `= (tan\ n theta-tan (n + 1) theta)/(tan (n theta-(n + 1) theta)`
  `= (tan\ n theta-tan (n + 1) theta)/(tan (-theta))`
  `=1/tan theta *-(tan\ n theta-tan (n + 1) theta)`
  `= cot theta (tan (n + 1) theta-tan\ n theta)`
  `=\ text(RHS  …  as required.)`

 

b.   `text(Prove)\ \ tan theta\ tan 2 theta + tan 2 theta\ tan 3 theta + … + tan\ n theta \ tan (n + 1) theta`

`= -(n + 1) + cot theta\ tan (n + 1) theta`

`text(If)\ \ n = 1:`

`text(LHS)` `= tan theta\ tan 2 theta`
  `= cot theta (tan 2 theta-tan theta)-1\ \ \ text{(from (i))}`
  `= cot theta\ tan 2 theta-1-1`
  `= -2 + cot theta\ tan 2 theta`
`text(RHS)` `= -2 + cot theta\ tan 2 theta`

 
`:.\ text(True for)\ \ n = 1`
 

`text(Assume true for)\ \ n = k:`

`text(i.e.)\ \ tan theta\ tan 2 theta + … + tan\ k theta\ tan (k + 1) theta`

`= -(k + 1) + cot theta\ tan (k + 1) theta`
 

`text(Prove true for)\ \ n = k + 1:`

`text(i.e.)\ \ tan theta\ tan 2 theta + … + tan\ k theta\ tan (k + 1) theta + tan (k + 1) theta\ tan (k + 2) theta`

`= -(k + 2) + cot theta\ tan (k + 2) theta`

`text(LHS)` `= -(k + 1) + cot theta\ tan(k + 1) theta + tan (k + 1) theta\ tan (k + 2) theta`
  `= -(k + 1) + cot theta\ tan(k + 1) theta + cot theta[tan (k + 2) theta-tan (k + 1) theta]-1`
  `= -(k + 2) + cot theta\ tan(k + 1) theta + cot theta\ tan (k + 2) theta-cot theta\ tan (k + 1) theta`
  `= -(k + 2) + cot theta\ tan(k + 2) theta`
  `=\ text(RHS)`

 
`=>\ text(True for)\ \ n = k + 1`

`:.\ text(S)text(ince true for)\ \ n = 1,\ \ text(true for integral)\ \ n >= 1`

Proof, EXT2* P2 2005 HSC 4d

Use the principle of mathematical induction to show that  `4^n - 1 - 7n > 0`  for all integers  `n >= 2.`  (3 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

`text(Prove)\ \ 4^n – 1 – 7n > 0\ \ text(for)\ \ n >= 2`

`text(If)\ \ n = 2`

`text(LHS)` `= 4^2 – 1 – (7 xx 2)`
  `= 16 – 1 – 14`
  `= 1 > 0`

 
`:.\ text(True for)\ \ n = 2`

 
`text(Assume true for)\ \ n = k`

`text(i.e.)\ \ 4^k – 1 – 7k` `> 0`
`4^k` `> 1 + 7k`

 
`text(Prove true for)\ \ n = k + 1`

`text(i.e.)\ \ 4^(k + 1) – 1 – 7 (k + 1) > 0`

`text(LHS)` `= 4 (4^k) – 1 – 7k – 7`
  `> 4 (1 + 7k) – 7k – 8`
  `> 4 + 28k – 7k -8`
  `> 21k – 4`
  `> 0\ \ \ \ (k >= 2)`

 
`=>\ text(True for)\ \ n = k + 1`

 `:.\ text(S)text(ince true for)\ \ n = 2, text(by PMI, true for integral)\ \ n >= 2.`

Proof, EXT2 P2 EQ-Bank 28

Prove by mathematical induction that  `2^n > n^2`  for integral  `n > 4`.  (3 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution
IMPORTANT: Inequality induction is challenging. Students need to state clearly what they need to prove and then use logic to advance their answer toward the desired result.

`text(Prove)\ \ 2^n > n^2\ \ text(for integral)\ \ n > 4`

`text(If)\ \ n = 5,`

`text(LHS)` `= 2^5 = 32`
`text(RHS)` `= 5^2 = 25 < text(LHS)`

 
`:.\ text(True for)\ n = 5`

 
`text(Assume true for)\ n = k:`

`text(i.e.)\ \ \ 2^k > k^2`

`text(Prove true for)\ n = k + 1`

`text(i.e.)\ \ \ ` `2^(k + 1)` `> (k + 1)^2`
  `2^(k + 1)` `> k^2 + 2k + 1`

 

`text(LHS)` `= 2^(k + 1)`
  `= 2*2^k`
  `> 2k^2`
  `> k^2 + k^2`
  `> k^2 + 4k\ \ text{(} text(noting)\ k^2>4k,\ k>4 text{)}`
  `> k^2 + 2k + 8\ \ text{(} text(noting)\ 4k > 2k + 8,\ k > 4 text{)}`
  `> k^2 + 2k + 1`
  `> (k + 1)^2`

 

`=>\ text(True for)\ n = k + 1`

`:.\ text(S)text(ince true for)\ \ n = 5, text(by PMI, true for integral)\ \ n > 4.`

Proof, EXT2 P2 EQ-Bank 13

Use mathematical induction to prove that

`sum_(r=1)^n r^3 = 1/4 n^2 (n + 1)^2`

`text(for integers)\ n>=1`  (3 marks)

Show Answers Only

 `text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

`text(Need to prove)\  sum_(r=1)^n r^3 = 1/4 n^2 (n + 1)^2\ \ text(integral)\ n>=1 `

`text(i.e.)\ 1^3 + 2^3 + 3^3 + … + n^3 = 1/4 n^2 (n + 1)^2`

`text(When)\ n = 1`

`text(LHS) = 1^3 = 1`

`text(RHS) = 1/4 1^2 (1 + 1)^2 = 1`

`:.\ text(True for)\ n = 1`

 
`text(Assume true for)\ n = k`

`text(i.e.)\ 1^3 + 2^3 + … + k^3 = 1/4 k^2 (k + 1)^2`

`text(Need to prove true for)\ n = k + 1`

`1^3 + 2^3 + … + k^3 + (k + 1)^3 = 1/4 (k + 1)^2 (k + 2)^2`

`text(LHS)` `= 1/4 k^2 (k + 1)^2 + (k + 1)^3`
  `= 1/4 (k + 1)^2 [k^2 + 4(k + 1)]`
  `= 1/4 (k + 1)^2 (k^2 + 4k + 4)`
  `= 1/4 (k + 1)^2 (k + 2)^2`
  `=\ text(RHS)`

 
`=>text(True for)\ n = k + 1`

`:.text(S)text(ince true for)\ n = 1,\ text(true for integral)\ n >= 1`

Proof, EXT2* P2 2009 HSC 7a

  1. Use differentiation from first principles to show that  `d/(dx)(x)=1`.   (1 mark)

  2. Use mathematical induction and the product rule for differentiation to prove that  
     
        `d/(dx)(x^n)=nx^(n-1)`  for all positive integers `n`.   (2 marks)

Show Answer Only

a.    `text{Proof (See Worked Solutions)}`

b.    `text{Proof (See Worked Solutions)}`

Show Worked Solution

a.    `text(Prove)\ \ d/(dx)(x)=1\ \ text(from first principles.)`

MARKER’S COMMENT: Students are reminded to use the number of marks allocated to a question as a guide to the work involved in answering it.
`d/(dx)(x)` `=lim_(h->0) (f(x+h)-f(x))/h`
  `=lim_(h->0)(x+h-x)/h`
  `=1\ \ \ text(… as required)`

 

b.    `text(Prove)\ \ d/(dx)(x^n)=nx^(n-1)\ \ text(for integers)\ n>=1`

`text(If)\  n=1,`

`text(LHS)=d/(dx)(x^1)=1`

`text(RHS)=(1)x^0=1=text(LHS)`

`:.\ text(True for)\ n=1`

 
`text(Assume true for)\ n=k`

`text(i.e.)\ \ d/(dx)(x^k)=kx^(k-1)`

 `text(Prove true for)\ n=k+1`

IMPORTANT: Critical to read carefully and apply the product rule to prove the induction.

`text(i.e.)\ \ d/(dx)x^(k+1)=(k+1)x^k`

`text(LHS)` `=d/(dx)(x^(k+1))`
  `=d/(dx) (x*x^k)`
  `=d/(dx)(x) * x^k+x *d/(dx)(x^k)`
  `=(1xx x^k)+(x xxkx^(k-1))`
  `=x^k+kx^k`
  `=(k+1)x^k`
  `=\ text(RHS … as required)`

 
`=>\ text(True for)\ n=k+1`

`:.\ text(S)text(ince it is true for)\ n=1, text(by PMI, true for integral)\ n>=1`.

Proof, EXT2* P2 2013 HSC 14a

  1. Show that for  `k>0,\ \ 1/(k+1)^2-1/k+1/(k+1)<0`.    (1 mark)

  2. Use mathematical induction to prove that for all integers  `n>=2`,
  3.    `1/1^2+1/2^2+1/3^2+\ …\ +1/n^2<2-1/n`.   (3 marks)

Show Answer Only

a.    `text{Proof (See Worked Solutions)}`

b.    `text{Proof (See Worked Solution)}`

Show Worked Solution

a.  `text(Prove that for)\ k>0,\ \ 1/(k+1)^2-1/k+1/(k+1)<0`

♦♦ Mean mark (a) 25%.
MARKER’S COMMENT: Creating a common denominator and justifying why the final expression is always negative caused issues.
`text(LHS)` `=(k-(k+1)^2+k(k+1))/(k(k+1)^2)`
  `=(k-(k^2+2k+1)+k^2+k)/(k(k+1)^2)`
  `=(-1)/(k(k+1)^2)`

 
`text(S)text(ince)\ \ k>0,\ text(we know)\ \ k(k+1)^2>0`

`:.\ (-1)/(k(k+1)^2)<0\ \ text(… as required)`

 

b.  `text(Prove)\ \ 1/1^2+1/2^2+1/3^2+\ …\ +1/n^2<2-1/n\ \ text(for integer)\ \ n>=2`

`text(If)\ \ n=2:`

`text(LHS)=1/1^2+1/2^2=5/4`

`text(RHS)=2-1/2=3/2 > text(LHS)`

`:.\ text(True for)\ \ n=2`

 
`text(Assume true for)\ \ n=k:`

`text(i.e.)\ \ 1/1^2+1/2^2+\ …\ +1/k^2<2-1/k`

 
`text(Prove true for)\ \ n=k+1:`

`text(i.e.)\ 1/1^2+1/2^2+\ …\ +1/k^2+1/(k+1)^2<2-1/(k+1)`

♦♦ Mean mark (b) 26%.
IMPORTANT: Students have to carefully examine part (i) of this question to provide the information required to prove true for `n=k+1`.
`text(LHS)` `=1/1^2+1/2^2+\ …\ +1/k^2+1/(k+1)^2`
  `<2-1/k+1/(k+1)^2`
  `< underbrace{(1/(k+1)^2-1/k+1/(k+1))}_text(<0  from part i)+2-1/(k+1)`
  `<2-1/(k+1)`

  
`=> text(True for)\ n=k+1`

`:.\ text(S)text(ince it is true for)\ n=2, text(by PMI, true for integral)\  n>=2`.