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

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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 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 2021 HSC 12d

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

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`text{See Worked Solutions}`

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`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)

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`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 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)

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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 2011 HSC 3c

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

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`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 2005 HSC 4d

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

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`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)

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`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 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)

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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`.