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Proof, EXT1 EQ-Bank 16

Consider the following proposition:

\(1+2+3+\ldots+n=\dfrac{1}{2}(n-1)(n+2)\)  for integers  \(n \geqslant 1\).

  1. Show that the proposition does not satisfy the initial case of mathematical induction proof.   (1 mark)

  2. Show that the inductive step of the false proposition can, however, be proven.   (2 marks)

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a.    \(\text{Initial case:} \ n=1\)

\(\text{LHS}=1\)

\(\text{RHS}=\dfrac{1}{2}(1-1)(1+2)=0 \neq \text{LHS}\)

\(\therefore \ \text{Initial case of} \ \ n=1\ \ \text {is not satisfied.}\)
 

b.    \(\text{if} \ \ n=k:\)

\(1+2+\ldots+k=\dfrac{1}{2}(k-1)(k+2)\)
 

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

\(\text{i.e.} \ \  1+2+\ldots+k+(k+1)=\dfrac{1}{2} k(k+3)\)

\(\operatorname{LHS}\) \(=1+2+\ldots+k+k+1\)
  \(=\dfrac{1}{2}(k-1)(k+2)+k+1\)
  \(=\dfrac{1}{2}\left(k^2+k-2\right)+\dfrac{1}{2}(2 k+2)\)
  \(=\dfrac{1}{2}\left(k^2+k-2+2 k+2\right)\)
  \(=\dfrac{1}{2}\left(k^2+3 k\right)\)
  \(=\dfrac{1}{2} k(k+3)\)
  \(=\operatorname{LHS}\)

 

\(\therefore \ \text{The inductive step can be proven.}\)

Proof, EXT1 EQ-Bank 20

Use mathematical induction to prove that  \(3^n+7^n\)  is divisible by 10 for all odd \(n\).   (3 marks)

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

Show Worked Solution

\(\text{Prove \(3^n+7^n\) is divisible by 10 for \(n\) odd.}\)

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

\(3^1+7^1=10 \ \ \text {(which is divisible by 10)}\)

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

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

\(3^k+7^k=10 P \ \text{(where \(P\) is an integer)}\)

\(\Rightarrow 3^k=10 P-7^k\ \ldots\ (1)\)
 

\(\text{Prove true for}\ \ n=k+2:\)

\(3^{k+2}+7^{k+2}\) \(=9 \times 3^k+49 \times 7^k\)
  \(=9\left(10 P-7^k\right)+49 \times 7^k\)
  \(=90 P-9 \times 7^k+49 \times 7^k\)
  \(=90P+40 \times 7^k\)
  \(=10\left(9 P + 4 \times 7^k\right)\)

 

\(\Rightarrow \ \text{True for} \ \ n=k+2\)

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

Proof, EXT1 P1 2025 SPEC1 7

Use mathematical induction to prove that

\begin{align*}
\displaystyle \sum_{i=1}^n(i+1)^2=\frac{1}{6} n\left(2 n^2+9 n+13\right) \text { for all integers}\ n\geq 1,
\end{align*}

where  \(\displaystyle \sum_{i=1}^n(i+1)^2=2^2+3^2+4^2+\ldots+(n+1)^2\).   (4 marks)

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\(\text{See Worked Solutions}\)

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\(\text{Prove} \ \ \displaystyle \sum_{i=1}^n(i+1)^2=\dfrac{1}{6} n\left(2 n^2+9 n+13\right)\)

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

\(\text{LHS}=(1+1)^2=4\)

\(\text {RHS}=\dfrac{1}{6} \times 1(2+9+13)=4\)

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

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

\(\text{i.e.} \ \ \displaystyle \sum_{i=1}^k(i+1)^2=\frac{1}{6} k\left(2 k^2+9 k+13\right)\ \ldots\ (1)\)

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

\(\text{i.e.} \ \ \displaystyle \sum_{i=1}^{k+1}(i+1)^2\) \(=\dfrac{1}{6}(k+1)\left[2(k+1)^2+9(k+1)+13\right]\)
  \(=\dfrac{1}{6}(k+1)\left(2 k^2+4 k+2+9 k+9+13\right)\)
  \(=\dfrac{1}{6}(k+1)\left(2 k^2+13 k+24\right)\)

 

\(\operatorname{LHS}\) \(=\dfrac{1}{6} k\left(2 k^2+9 k+13\right)+(k+2)^2\ \ \text{(using (1) above)}\)
  \(=\dfrac{1}{6}\left(2 k^3+9 k^2+13 k\right)+\left(k^2+4 k+4\right)\)
  \(=\dfrac{1}{6}\left(2 k^3+9 k^2+13 k+6 k^2+24 k+24\right)\)
  \(=\dfrac{1}{6}\left(2 k^3+15 k^2+37 k+24\right) \quad \text{(see long division below)}\)
  \(=\dfrac{1}{6}(k+1)\left(2 k^2+13 k+24\right)\)
  \(=\operatorname{RHS}\)

 

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

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

Proof, EXT1 P1 2025 HSC 12c

Prove by mathematical induction that

\(1 \times(1!)+2 \times(2!)+\cdots+n \times(n!)=(n+1)!-1\)

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

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\(\text{See Worked Solution}\)

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\(1 \times 1!+2 \times 2!+\ldots+n \times n!=(n+1)!-1\)

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

\(\text{LHS}=1 \times 1!=1\)

\(\text{RHS}=(2)!-1=1=\text{LHS }\)

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

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

\(1 \times 1!+2 \times 2!+\ldots+k \times k!=(k+1)!-1\)
 

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

\(\text{i.e.} \ \ 1 \times 1!+2 \times 2!+\ldots +k \times k!+(k+1) \times(k+1)!=(k+2)!-1\)

\(\text{LHS}\) \(=1\times 1!+\ldots+k \times k!+(k+1) \times(k+1)!\)
  \(=(k+1)!-1+(k+1) \times (k+1)!\)
  \(=(k+1)!(1+k+1)-1\)
  \(=(k+1)!(k+2)-1\)
  \(=(k+2)!-1\)
  \(= \operatorname{RHS}\)

 

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

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

Proof, EXT1 P1 EQ-Bank 32

  1. Show that  \(\cos (A-B)-\cos (A+B)=2 \sin A \sin B\).   (1 mark)

  2. Using the result in part (a), or otherwise, prove by mathematical induction that, for \(n \geqslant 1\)
  3. \(\sin \theta+\sin (3 \theta)+\ldots+\sin ((2 n-1) \theta)=\dfrac{\sin ^2(n \theta)}{\sin \theta}\)   (3 marks)
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a.     \(\cos (A-B)-\cos (A+B)\) \(=\cos A \cos B+\sin A \sin B-(\cos A \cos B-\sin A \sin B)\)
    \(=2 \sin A \cos B\).

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

Show Worked Solution
a.     \(\cos (A-B)-\cos (A+B)\) \(=\cos A \cos B+\sin A \sin B-(\cos A \cos B-\sin A \sin B)\)
    \(=2 \sin A \cos B\).

 

b.    \(\text{RTP:} \ \ \sin \theta+\sin (3 \theta)+\ldots+\sin ((2 n-1) \theta)=\dfrac{\sin ^2(n \theta)}{\sin \theta}\)

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

\(\text{LHS}=\sin \theta\)

\(\text {RHS}=\dfrac{\sin ^2 \theta}{\sin \theta}=\sin \theta=\text{LHS}\)

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

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

\(\sin \theta+\sin (3 \theta)+\ldots+\sin \left((2 k-1) \theta\right)=\dfrac{\sin ^2(k \theta)}{\sin \theta}\ \ …\ \text{(*)}\)
 

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

\(\sin \theta+\sin (3 \theta)+\cdots+\sin ((2 k-1) \theta)+\sin ((2 k+1) \theta)=\dfrac{\sin ^2((k+1) \theta)}{\sin \theta}\)

\(\text {LHS}\) \(=\dfrac{\sin ^2(k \theta)}{\sin \theta}+\sin ((2 k+1) \theta) \quad \text{(see (*) above)}\)
  \(=\dfrac{\sin ^2(k \theta)+\sin \theta \times \sin ((2 k+1) \theta)}{\sin \theta}\ \ …\ (1)\)

 
\(\text{Using part a:}\)

\(\sin \theta \cdot \sin ((2 k+1) \theta)\) \(=\dfrac{1}{2}[\cos (2 k \theta)-\cos ((2 k+2) \theta)]\)
  \(=\dfrac{1}{2}[\cos (2 k \theta)-\cos (2(k+1) \theta)]\)
  \(=\dfrac{1}{2}\left[1-2 \sin ^2(k \theta)-\left(1-2 \sin ^2((k+1) \theta)\right)\right]\)
  \(=\dfrac{1}{2}\left(2 \sin ^2((k+1) \theta)-2 \sin ^2(k \theta)\right)\)
  \(=\sin ^2((k+1) \theta)-\sin ^2(k \theta)\)

 
\(\text{Substitute into (1):}\)

\(\text {LHS}\) \(=\dfrac{\sin ^2(k \theta)+\sin ^2((k+1) \theta)-\sin ^2(k \theta)}{\sin \theta}\)
  \(=\dfrac{\sin ^2((k+1) \theta)}{\sin \theta}=\text{RHS}\)

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

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

Proof, EXT1 P1 2024 HSC 12d

Use mathematical induction to prove that  \(2^{3 n}+13\)  is divisible by 7 for all integers  \(n \geq 1\).   (3 marks)

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

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\(\text{Prove}\ \ 2^{3 n}+13\ \ \text{is divisible by 7 for}\ \ n \geq 1.\)

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

\(2^{3 \times 1}+13=21=3 \times 7\)

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

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

\(2^{3 k}+13=7P \ \text{(where \(P\) is an integer)}\)

\(\Rightarrow 2^{3k}=7 P-13\ \ldots\ (1)\)

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

  \(2^{3(k+1)}+13\) \(=2^{3 k} \times 2^3+13\)
    \(=8\left(2^{3 k}\right)+13\)
    \(=8(7P-13)+13\ \ \text{(see (1) above)}\)
    \(=56 P-8 \times 13+13\)
    \(=7(8 P-13)\)

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

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

Proof, EXT1 P1 2023 HSC 12b

Use mathematical induction to prove that

\((1 \times 2)+\left(2 \times 2^2\right)+\left(3 \times 2^3\right)+\cdots+\left(n \times 2^n\right)=2+(n-1) 2^{n+1}\)

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

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

Show Worked Solution

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

\(\text{LHS}\ = 1 \times 2=2 \)

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

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

\((1 \times 2)+\left(2 \times 2^2\right)+\cdots+\left(k \times 2^k\right)=2+(k-1) 2^{k+1}\)
 

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

\((1 \times 2)+\left(2 \times 2^2\right)+\cdots+\left(k \times 2^k\right) + (k+1)2^{k+1}=2+k \times 2^{k+2}\)

\(\text{LHS}\) \(=2+(k-1)2^{k+1} + (k+1) 2^{k+1} \)  
  \(=2+2^{k+1}(k-1+k+1) \)  
  \(=2+2^{k+1}(2k) \)  
  \(=2+k \times 2^{k+2} \)  
  \(=\ \text{RHS} \)  
 

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

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

Proof, EXT1 P1 2022 HSC 12f

Use mathematical induction to prove that  `15 ^(n)+6^(2n+1)`  is divisible by 7 for all integers  `n >= 0`.  (3 marks)

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`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text{Prove}\ \ 15 ^(n)+6^(2n+1)\ \ text{is divisible by 7 for}\ \ n>=0`

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

`15^0+6^1=7`

`:.\ text{True for}\ \ n=0`
 

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

`text{i.e.}\ \ 15^(k)+6^(2k+1)=7P\ \ text{(where}\ P\ text{is an integer)}`

`=>6^(2k+1)=7P-15^(k)\ \ …\ (1)`
 

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

`15^(k+1)+6^(2k+3)` `=15*15^(k)+6^2*6^(2k+1)`  
  `=15*15^(k)+36(7P-15^(k))\ \ text{(see (1))}`  
  `=15*15^(k)+36*7P-36*15^(k)`  
  `=36*7P-21*15^(k)`  
  `=7(36P-3*15^(k))\ \ text{(which is divisible by 7)}`  

 
`=>\ text{True for}\ \ n=k+1`

`:.\ text{S}text{ince true for}\ \ n=0, text{by PMI, true for integers}\ \ n>=0`

Proof, EXT1 P1 2021 HSC 12c

Use mathematical induction to prove that

`1/(1 xx 2 xx 3) + 1/(2 xx 3 xx 4) + … + 1/(n(n + 1)(n + 2)) = 1/4 - 1/(2(n + 1)(n + 2))`

 
for all integers `n ≥ 1`.  (3 marks)

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`text(See Worked Solutions)`

Show Worked Solution

`text(Prove for)\ \ n >= 1,`

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

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

`text(LHS)\ = 1/(1 xx 2 xx 3) = 1/6`

`text(RHS)\ = 1/4 – 1/(2 xx 2 xx 3) = 1/4 – 1/12 = 1/6`

 
`:.\ text(True for)\ \ n = 1`
 

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

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

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

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

`text(LHS)` `= 1/(1 xx 2 xx 3) + … + 1/(k(k + 1)(k + 2)) + 1/((k + 1)(k + 2)(k + 3))`
  `= 1/4 – 1/(2(k + 1)(k + 2)) + 1/((k + 1)(k + 2)(k + 3))`
  `= 1/4 – ((k + 3 – 2)/(2(k + 1)(k + 2)(k + 3)))`
  `= 1/4 – (k + 1)/(2(k + 1)(k + 2)(k + 3))`
  `= 1/4 – 1/(2(k + 2)(k + 3))`
  `=\ 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, EXT1 P1 2020 HSC 12a

Use the principle of mathematical induction to show that for all integers  `n >= 1`,

`1 xx 2 + 2 xx 5 + 3 xx 8 + … + n(3n-1) = n^2(n + 1)`.  (3 marks)

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`text(See Worked Solutions)`

Show Worked Solution

`text(Prove)`

`1 xx 2 + 2 xx 5 + 3 xx 8 + … + n(3n-1) = n^2(n + 1)`
 

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

`text(LHS) = 1(3 xx 1-1) = 2`

`text(RHS) = 1^2(1 + 1) = 2`
 

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

`text(i.e.)\ \ 1 xx 2 + 2 xx 5 + 3 xx 8 + … + k(3k-1) = k^2(k + 1)`
 

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

`text(i.e.)\ 1 xx 2 + 2 xx 5 + … + k(3k-1) + (k + 1)(3k + 2) = (k + 1)^2(k + 2)`

`text(LHS)` `= 1 xx 2 + … + k(3k-1) + (k + 1)(3k + 2)`
  `= k^2(k + 1) + (k + 1)(3k + 2)`
  `= (k + 1)(k^2 + 3k + 2)`
  `= (k + 1)(k + 1)(k + 2)`
  `= (k + 1)^2(k + 2)`
  `=\ 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, EXT1 P1 2019 HSC 14a

Prove by mathematical induction that, for all integers `n >= 1`,

`1(1!) + 2(2!) + 3(3!) + … + n(n!) = (n + 1)! - 1`.  (3 marks)

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

Show Worked Solution

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

`text(LHS) = 1(1!) = 1`

`text(RHS) = (1 + 1)! – 1 = 2! – 1 = 1 = text(LHS)`

`:.\ text(True for)\ \ n = 1`
 

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

`1(1!) + 2(2!) + … + k(k!) = (k + 1)! – 1`
 

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

`text(i.e.)\ \ underbrace(1(1!) + 2(2!) + … + k(k!))_((k + 1)! – 1) + (k + 1)(k + 1)! = (k + 2)! – 1`

`text(LHS)` `= (k + 1)! – 1 + (k + 1)(k + 1)!`
  `= (k + 1)! [1 + (k + 1)] – 1`
  `= (k + 1)!(k + 2) – 1`
  `= (k + 2)! – 1`

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

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

Proof, EXT1 P1 2018 HSC 13a

Prove by mathematical induction that, for  `n >= 1,`

`2-6 + 18-54 + … + 2 (-3)^(n-1) = (1-(-3)^n)/2.`  (3 marks)

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

Show Worked Solution

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

`text(LHS)` `= 2 (-3)^0 = 2`
`text(RHS)` `= (1-(-3)^1)/2 = 2`

`:.\ text(True for)\ n = 1`
 

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

`2-6 + 18-54 + … + 2 (-3)^(k-1) = (1-(-3)^k)`
 

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

`text(i.e.)\ \ 2-6 + … + 2 (-3)^(k-1) + 2 (-3)^k = (1-(-3)^(k + 1))/2`

`text(LHS)` `= underbrace{2-6 + … + 2(-3)^(k-1)}_text(Sum of GP, r = –3) + 2 (-3)^k`
  `= (2(1-(-3)^k))/(1+3) + 2 (-3)^k`
  `= (1-(-3)^k + 4 (-3)^k)/2`
  `= (1 + 3(-3)^k)/2`
  `= (1-(-3)(-3)^k)/2`
  `= (1-(-3)^(k + 1))/2`
  `=\ 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, EXT1 P1 2017 HSC 14a

Prove by mathematical induction that  `8^(2n + 1) + 6^(2n − 1)`  is divisible by 7, for any integer  `n ≥ 1`.  (3 marks)

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`text(See Worked Solutions)`

Show Worked Solution

`text(Prove)\ \ 8^(2n + 1) + 6^(2n − 1)\ \ text(is divisible)`

`text(by 7 for integers)\ \ n >= 1`

`text(If)\ n = 1,`

`8^3 + 6^1 = 518 = 74 xx 7`

`:. text(True for)\ n = 1`
 

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

`text(i.e.)\ \ 8^(2k + 1) + 6^(2k – 1)` `= 7P\ \ (text(where)\ P\ text(is an integer))`
`8^(2k + 1)` `= 7P – 6^(2k – 1)`

 
`text(Prove true for)\ n = k + 1`

`8^(2k + 3) + 6^(2k − 1)` `= 64 · 8^(2k + 1) + 36*6^(2k – 1)`
  `= 64(7P – 6^(2k – 1)) + 36 · 6^(2k – 1)`
  `= 64 · 7P – 64 · 6^(2k – 1) + 36 · 6^(2k – 1)`
  `= 64 · 7P – 28 · 6^(2k – 1)`
  `= 7(64P – 4 · 6^(2k – 1))`

 
`…\ text(which is divisible by 7.)`
 

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

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

Proof, EXT1 P1 2016 HSC 14a

  1. Show that  `4n^3 + 18n^2 + 23n + 9`  can be written as
  2. `qquad (n + 1)(4n^2 + 14n + 9)`.   (1 marks)

  3. Using the result in part (i), or otherwise, prove by mathematical induction that, for  `n >= 1`,
  4. `qquad 1 × 3 + 3 × 5 + 5 × 7 + … + (2n-1)(2n + 1) = 1/3 n(4n^2 + 6n-1)`.   (3 marks)
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a.    `text(Proof)\ \ text{(See Worked Solutions)}`

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

Show Worked Solution
a.     `text(RHS)` `= (n + 1) (4n^2 + 14n + 9)`
    `= 4n^3 + 14n^2 + 9n + 4n^2 + 14n + 9`
    `= 4n^3 + 18n^2 + 23n + 9`

 

b.    `text(Prove)\ \ 1 xx 3 + 3 xx 5 + 5 xx 7 + … + (2n-1) (2n + 1)`

`= 1/3n (4n^2 + 6n-1)`

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

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

`text(RHS) = 1/3 (1) (4 + 6-1) = 3`

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

 

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

`text(i.e.)\ \ 1 xx 3 + 3 xx 5 + … + (2k-1) (2k + 1)`

`= 1/3 k (4k^2 + 6k-1)`

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

`text(i.e.)\ \ 1 xx 3 + … + (2k + 1) (2k + 3)`

`= 1/3 (k + 1) [4 (k + 1)^2 + 6(k + 1)-1]`

`= 1/3 (k + 1) (4k^2 + 14k + 9)`

`= 1/3 (4k^3 + 18k^2 + 23k + 9)`

 

`text(LHS)` `= 1 xx 3 + … + (2k-1) (2k + 1) + (2k + 1) (2k + 3)`
  `= 1/3k (4k^2 + 6k-1) + (2k + 1) (2k + 3)`
  `= 1/3 (4k^3 + 6k^2-k) + (4k^2 + 8k + 3)`
  `= 1/3 (4k^3 + 6k^2-k + 12k^2 + 24k + 9)`
  `= 1/3 (4k^3 + 18k^2 + 23k + 9)`
  `=\ 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, EXT1 P1 2007 HSC 4b

Use mathematical induction to prove that  `7^(2n – 1) + 5`  is divisible by 12, for all integers  `n ≥ 1`.  (3 marks)

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`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text(Prove)\ \ 7^(2n – 1) + 5\ \ text(is divisible by)\ 12,\ text(for)\ \ n ≥ 1`

`text(If)\ \ n = 1`

`7^(2 – 1) + 5 = 7 + 5 = 12\ \ \ text{(divisible by 12)}`

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

 

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

`text(i.e.)\ 7^(2k − 1) + 5` `= 12text{N   (N is an integer)}`
 `7^(2k-1)` `= 12text(N) – 5\ \ …\ (1)`

 

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

`7^(2(k + 1) − 1) + 5` `= 7^(2k + 1) + 5`
  `= 7^2 · 7^(2k − 1) + 5`
  `= 49(12text(N) − 5) + 5\ \ \ text{(from (1) above)}`
  `= 49*12text(N) − 245 + 5`
  `= 49*12text(N) − 240`
  `= 12(49text(N) − 20)`

`…text(which is divisible by)\ 12`

 
`=>\ text(True for)\ n = k + 1`

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

Proof, EXT1 P1 2015 HSC 13c

Prove by mathematical induction that for all integers `n ≥ 1`,

`1/(2!) + 2/(3!) + 3/(4!) + … + n/((n + 1)!) = 1 − 1/((n + 1)!)`.  (3 marks)

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`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text(Prove for)\ n ≥ 1`

`1/(2!) + 2/(3!) + 3/(4!) + … + n/((n + 1)!) = 1 − 1/((n + 1)!)`

`text(If)\ n = 1`

`text(LHS)` `= 1/(2!) = 1/2`
`text(RHS)` `= 1 − 1/(2!) = 1 − 1/2 = 1/2`

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

 
`text(Assume true for)\ n = k`

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

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

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

`text(LHS)` `= 1 − 1/((k + 1)!) + (k + 1)/((k + 2)!)`
  `= 1 − (((k + 2) − (k + 1))/((k + 1)!(k + 2)))`
  `= 1 − 1/((k + 2)!)\ \ …\ text(as required)`

 
`=>\ text(True for)\ n = k + 1`

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

Proof, EXT1 P1 2008 HSC 3b

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

`1 xx 3 + 2 xx 4 + 3 xx 5 + ... + n(n+2) = n/6 (n + 1)(2n + 7)`.   (3 marks)

Show Answers Only

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

Show Worked Solution

`text(Prove)\ \ 1 xx 3 + 2 xx 4 + 3 xx 5 + … + n(n+2)`

`= n/6 (n+1)(2n + 7)\ \ \ text(for)\ n >= 1`

`text(If)\ \ n = 1`

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

`text(RHS) = 1/6 (2)(9) = 3 = text(LHS)`

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

 
`text(Assume true for)\ \ n = k`

`text(i.e.)\ \ 1 xx 3 + 2 xx 4 + … + k (k + 2)`

`= k/6 (k + 1)(2k + 7)`

 
`text(Prove true for)\ \ n = k + 1`

`text(i.e.)\ \ 1 xx 3 + 2 xx 4 + … + k(k+2) + (k + 1)(k + 3)`

`= ((k+1))/6 (k + 2)(2k + 9)`

`text(LHS)` `= k/6 (k + 1)(2k + 7) + (k + 1)(k + 3)`
  `= 1/6 (k + 1)[k(2k + 7) + 6(k + 3)]`
  `= 1/6 (k + 1)[2k^2 + 7k + 6k + 18]`
  `= 1/6 (k + 1)[2k^2 + 13k + 18]`
  `= 1/6 (k +1)(k + 2)(2k + 9)\ \ \ \ text(… as required)`

 
`=>\ text(True for)\ n = k + 1`

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

Proof, EXT1 P1 EQ-Bank 29

Prove by mathematical induction that  `(3n + 1)7^n -1`  is divisible by 9 for integral  `n >= 1`.   (3 marks)

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

Show Worked Solution

`text(Prove)\ \ (3n + 1)7^n-1\ \ text(is divisible)`

`text(by 9 for integers)\ n >= 1.`

`text(If)\ \ n = 1`

`(3n + 1)7^n-1` `= (3 xx 1 + 1)7^1-1`
  `= 27`

`=>\ text(Divisible by 9)\ \ (27//9=3)`
 

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

IMPORTANT: In cases where substitution doesn’t work, remember that `f(k+1)-f(k)` can be used to illustrate divisibility (Tony Lantry special).

`text(i.e.)\ (3k + 1)7^k-1 = 9P\ \ \ text{(} P\ text(integer) text{)}`

`f(k)` `= (3k + 1)7^k-1`
`f(k + 1)` `= (3 (k + 1) + 1)7^(k+1)-1`
  `= (3k + 4)7^(k+1)-1`

 
`f(k + 1)-f(k)`

`= (3k + 4) 7^(k + 1)-1- [(3k + 1)7^k-1]`
`= 3k * 7^(k + 1) + 4 * 7^(k + 1)-1-[3k*7^k + 7^k-1]`
`= 21k*7^k + 28*7^k-1-3k*7^k-7^k + 1`
`= 18k*7^k + 27*7^k`
`= 9 (2k*7^k + 3*7^k)\ \ text{(divisible)}`
 
`text(S)text(ince)\ \ ` `f(k + 1)-f(k)\ text(is divisible by 9 and)`
  `\ \ \ \ \ \ f(k)\ text(is divisible by 9,)`

 
`=> f(k + 1)\ text(is divisible by 9)`

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

Proof, EXT2 P1 EQ-Bank 28

Use mathematical induction to prove that  `7^n + 15^n`  is divisible by 11 where `n` is an odd integer.  (3 marks)

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

Show Worked Solution

`text(Prove)\ \ 7^n + 15^n\ \ text(is divisible by 11, where)\ n\ text(is odd)`

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

`7^n + 15^n` `= 7 + 15 = 22`
`22/11` `= 2\ \ \ text{(Divisible by 11)}`

 
`text(Assume true for)\ \ n = k:`

`text(i.e.)\ 7^k + 15^k` `= 11P\ \ \ \ text{(}P\ text(integer) text{)}`
`7^k` `= 11P\ – 15^k\ \ \ \ \ …\ text{(∗)}`

 
`text(Prove true for)\ \ n = k + 2:`

`7^(k + 2) + 15^(k + 2)` `= 7^2 * 7^k + 15^2 * 15^k`
  `= 49 (11P-15^k) + 225 (15^k)\ \ \ text(… from)\ text{(∗)}`
  `= 49 * 11P-49 * 15^k + 225 (15^k)`
  `= 49 * 11P + 176 * 15^k`
  `= 11 (49P + 16 * 15^k)`

  
`=>text(True for)\ n = k + 2`

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

Proof, EXT1 P1 2014 HSC 13a

Use mathematical induction to prove that  `2^n + (− 1) ^(n + 1)`  is divisible by 3 for all integers  `n >= 1`.   (3 marks)

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

Show Worked Solution

`text(Prove)\ \ 2^n + (-1)^(n + 1)`

`text(is divisible by 3 for integers)\ n >= 1`

`text(If)\ n = 1`

`2^1 + (-1)^2` `= 2 + 1`
  `= 3\ \ text{(which is divisible by 3)}`

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

 
`text(Assume true for)\ \ n = k`

`text(i.e.)\ \ \ 2^k + (-1)^(k + 1)` `= 3P\ text{(} P\ text(integer) text{)}`
`2^k` `= 3P\ – (-1)^(k + 1)\ \ \ \ \ … \ text{(∗)}`

 
`text(Prove true for)\ \ n = k + 1`

`2^(k + 1) + (-1)^(k + 1 + 1)` `= 2*2^k + (-1)^(k + 2)`
  `= 2 (3P\ – (-1)^(k + 1)) + (-1)^(k + 2)\ \ text(… from)\ text{(∗)}`
  `= 6P\ – 2 (-1)^(k + 1)\ – 1 (-1)^(k + 1)`
  `= 6P\ – 3 (-1)^(k + 1)`
  `= 3 (2P\ – (-1)^(k + 1))`
`text(… which is divisible by 3)`

 
`=>\ text(True for)\ n = k + 1`

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

Proof, EXT1 P1 2010 HSC 7a

Prove by induction that 

`47^n + 53 xx 147^(n-1)`

is divisible by  `100`  for all integers  `n >= 1`.    (3 marks)

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

Show Worked Solution

`text(Prove)\ \ 47^n + 53 xx 147^(n – 1)\ \ text(is divisible)`

`text(by 100 for)\ n >= 1.`

`text(If)\ n = 1`

`47^1 + 53 xx 147^(1 – 1)` `= 47 + 53`
  `= 100\ \ text{(divisible by 100)}`

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

♦ Mean mark 51%.
IMPORTANT: Making `47^k` the subject helps the substitution in the proof for `n+1`. Students must state that `P` is an integer. NOTE: using `147^k=147xx147^(k-1)` develops proof for `n=k+1`.

 
`text(Assume true for)\ \ n = k`

`text(i.e.)\ 47^k + 53 xx 147^(k – 1) = 100P\ \ \ text{(} P\ text(integer) text{)}`

`47^k = 100P\ – (53 xx 147^(k – 1))\ \ \ text(…)\ text{(1)}`

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

`47^(k + 1) + 53 xx 147^(k)`
`= 47 xx 47^k + 53 xx 147^k`
`text{(substitute from (1) above)}`
`= 47 [100P\ – (53 xx 147^(k – 1))] + 53 xx 147 xx 147^(k – 1)`
`= 4700P\ – 2491 xx 147^(k – 1) + 7791 xx 147^(k – 1)`
`= 4700P + 5300 xx 147^(k – 1)`
`= 100 (47P + 53 xx 147^(k – 1))`

 
`=>\ text(True for)\ n = k + 1`

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

Proof, EXT1 P1 2012 HSC 12a

Use mathematical induction to prove that  `2^(3n)\ – 3^n`  is divisible by  `5`  for  `n >= 1`.   (3 marks)

Show Answers Only

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

Show Worked Solution

`text(Prove by induction)`

`2^(3n)\ – 3^n\ text(is divisible by)\ 5\ text(for)\ n >= 1`

`text(If)\ n = 1`

`2^(3n)\ – 3^n` `= 2^3\ – 3^1`
  `= 5\ \ \ \ text{(divisible by 5)}`

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

IMPORTANT: Making `3^k` or `2^(3k)` the subject helps for the substitution required in the proof for `n+1`. Important to state that `P` is an integer.

 
`text(Assume true for)\ n = k`

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

 
`text(Prove true for)\ n = k+1`

`2^(3(k+1))\ – 3^(k+1)` `= 2^(3k + 3)\ – 3 * 3^k`
  `= 2^3 * 2^(3k)\ – 3(2^(3k)\ – 5P)\ \ \ text{(from (1) above)}`
  `= 8 * 2^(3k)\ – 3 * 2 ^(3k) + 15P`
  `= 5 * 2^(3k) + 15P`
  `= 5 (2^(3k) + 3P)`

 
`=>\ text(True for)\ n = k + 1`

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

Proof, EXT1 P1 2011 HSC 6a

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

`1xx5+2xx6+3xx7+\ …\ +n(n+4)=1/6n(n+1)(2n+13)`.   (3 marks)

Show Answer Only

`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text(Prove)\ \ 1xx5+2xx6+3xx7+\ …\ +n(n+4)`

`=1/6n(n+1)(2n+13)\ text(for)\ n>=1`

`text(If)\ n=1`

`text(LHS)=1xx5=5`

`text(RHS)=1/6(1)(2)(15)=30/6=5=text(LHS)`

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

 
`text(Assume true for)\ n=k`

`text(i.e.)\ \ 1xx5+2xx6+3xx7+\ …\ +k(k+4)`

`=1/6k(k+1)(2k+13)`

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

`text(i.e.)\ 1xx5+2xx6+\ …\ +k(k+4)+(k+1)(k+5)`

`=1/6(k+1)(k+2)(2k+15)`

MARKER’S COMMENT: Write out the statement to be proven. Transcription errors, poor setting out and inefficient algebraic approaches (factorise whenever possible) were common errors.
`text(LHS)` `=1/6k(k+1)(2k+13)+(k+1)(k+5)`
  `=1/6(k+1)[k(2k+13)+6(k+5)]`
  `=1/6(k+1)[2k^2+13k+6k+30]`
  `=1/6k(k+1)(2k^2+19k+30)`
  `=1/6k(k+1)(k+2)(2k+15)`
  `=\ text(RHS … as required)`

 
`=>\ text(True for)\ n=k+1`

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