smc-1019-20-Sum of a Series

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

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

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

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

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

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

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

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

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

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

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

Proof, EXT1 P1 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)`

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`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)`

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`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)}`

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`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)}`

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`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 2016 HSC 14a

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

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Using the result in part (i), or otherwise, prove by mathematical induction that, for `n >= 1`,
`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 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)}`

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`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)

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

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`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 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)

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`text{Proof (See Worked Solutions)}`

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`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`.