2013 H2 Mathematics Paper 1 Question 9

Sigma Notation

Answers

(i)

Out of syllabus

(ii)

f(r)−f(r−1)=6r2{f(r)-f(r-1)=6r^2}
16n(n+1)(2n+1){\frac{1}{6}n(n+1)(2n+1)}

(iii)

12n(n+1)(n2+n+1)+12n(n+1)(2n+1)+24n{\frac{1}{2}n(n+1)(n^2+n+1)} \allowbreak {+ \frac{1}{2}n(n+1)(2n+1)} \allowbreak {+ 24n}

Full solutions

(i)

Out of syllabus

(ii)

f(r−1)=2(r−1)3+3(r−1)2+(r−1)+24=2(r3−3r2+3r−1)+3(r2−2r+1)+(r−1)+24=2r3−3r2+r+24\begin{align*} &f(r-1) \\ &= 2(r-1)^3 + 3(r-1)^2 + (r-1) + 24 \\ &= 2(r^3-3r^2+3r-1) + 3(r^2-2r+1) + (r-1) + 24 \\ &= 2 r^3 - 3 r^2 + r + 24 \\ \end{align*}
f(r)−f(r−1)=2r3+3r2+r+24−(2r3−3r2+r+24)=6r2  ■\begin{align*} &f(r)-f(r-1) \\ &=2 r^3 + 3 r^2 + r + 24-\left(2 r^3 - 3 r^2 + r + 24\right) \\ &= 6 r^2 \; \blacksquare \end{align*}
∑r=1nr2=∑r=1nf(r)−f(r−1)6=16(f(1)−f(0)+f(2)−f(1)+f(3)−f(2)⋯f(n−2)−f(n−3)f(n−1)−f(n−2)f(n)−f(n−1))=16(f(n)−f(0))=16(2n3+3n2+n+24−24)=16n(2n2+3n+1)=16n(n+1)(2n+1)  ■\begin{align*} &\sum_{r=1}^n r^2 \\ &= \sum_{r=1}^n \frac{f(r)-f(r-1)}{6} \\ & = \frac{1}{6}\left( \def\arraystretch{1.5} \begin{array}{lclc} & \bcancel{f(1)} &-& f(0) \\ + & \bcancel{f(2)} &-& \bcancel{f(1)} \\ + & \bcancel{f(3)} &-& \bcancel{f(2)} \\ & & \cdots & \\ & \bcancel{f(n-2)} &-& \bcancel{f(n-3)} \\ & \bcancel{f(n-1)} &-& \bcancel{f(n-2)} \\ & f(n) &-& \bcancel{f(n-1)} \\ \end{array} \right) \\ &= \frac{1}{6}\Big(f(n)-f(0)\Big) \\ &= \frac{1}{6}\Big(2 n^3 + 3 n^2 + n + 24-24\Big) \\ &= \frac{1}{6}n(2n^2+3n+1) \\ &= \frac{1}{6}n(n+1)(2n+1) \; \blacksquare \end{align*}

(iii)

Result from (i):
∑r=1nr(2r2+1)=12n(n+1)(n2+n+1)\begin{align} && \quad \sum_{r=1}^n r(2r^2+1) = \frac{1}{2}n(n+1)(n^2+n+1) \end{align}
∑r=1nf(r)=∑r=1n(2r3+3r2+r+24)=∑r=1nr(2r3+1)+3∑r=1nr2+∑r=1n24=12n(n+1)(n2+n+1)+12n(n+1)(2n+1)+24n  ■\begin{align*} &\sum_{r=1}^n f(r) \\ &= \sum_{r=1}^n \left( 2r^3 + 3r^2 + r + 24 \right) \\ &= \sum_{r=1}^n r(2r^3+1) + 3\sum_{r=1}^n r^2 + \sum_{r=1}^n 24 \\ &= \frac{1}{2}n(n+1)(n^2+n+1) + \frac{1}{2}n(n+1)(2n+1) + 24n \; \blacksquare \end{align*}