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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 EQ-Bank 9

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 7

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.`

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 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 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
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `text(See Worked Solutions)`
  4. `text(See Worked Solutions)`
  5. `text(See Worked Solutions)`
Show Worked Solution

i.   `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)`

 

ii.   `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%.
iii.    `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)]`

 

iv.   `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`

 

v.   `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 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
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `4/pi`
Show Worked Solution

i.   `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)`

 

ii.   `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.`

 

iii.   `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`

 

iv.   `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 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
  1. `text{Proof (See Worked Solutions.)}`
  2. `text{Proof (See Worked Solutions.)}`
  3. `pi/4`
Show Worked Solution

i.   `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)`

 

ii.  `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`

 

iii. `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 EQ-Bank 10

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`