2015 H2 Mathematics Paper 2 Question 4

Sigma Notation

Answers

(a)

Out of syllabus
(bi)
12r+1+−12r+3{\frac{1}{2 r + 1}+\frac{- 1}{2 r + 3}}
(bii)
Sn=13−12n+3{S_n = \frac{1}{3} - \frac{1}{2 n + 3}}
(biii)
Smallest value of n=499{n = 499 }

Full solutions

(a)

Out of syllabus
(bi)
By the cover up rule,
24r2+8r+3=2(2r+1)(2r+3)=12r+1+−12r+3  ■\begin{align*} &\frac{2}{4 r^2 + 8 r + 3} \\ &= \frac{2}{(2 r + 1)(2 r + 3)} \\ &= \frac{1}{2 r + 1} + \frac{- 1}{2 r + 3} \; \blacksquare \end{align*}
(bii)
Sn=∑r=1n24r2+8r+3=∑r=1n(12r+1−12r+3)=13−15+15−17+17−19+⋯+12n−3−12n−1+12n−1−12n+1+12n+1−12n+3=13−12n+3  ■\begin{align*} & S_n \\ &= \sum_{r=1}^n \frac{2}{4 r^2 + 8 r + 3} \\ &= \sum_{r=1}^n \left(\frac{1}{2 r + 1} - \frac{1}{2 r + 3}\right) \\ & = \def\arraystretch{1.5} \begin{array}{lclc} & \frac{1}{3} &-& \cancel{\frac{1}{5}} \\ + & \cancel{\frac{1}{5}} &-& \cancel{\frac{1}{7}} \\ + & \cancel{\frac{1}{7}} &-& \cancel{\frac{1}{9}} \\ + & & \cdots & \\ + & \cancel{\frac{1}{2 n - 3}} &-& \cancel{\frac{1}{2 n - 1}} \\ + & \cancel{\frac{1}{2 n - 1}} &-& \cancel{\frac{1}{2 n + 1}} \\ + & \cancel{\frac{1}{2 n + 1}} &-& \frac{1}{2 n + 3} \end{array} \\ \\ &= \frac{1}{3} - \frac{1}{2 n + 3} \; \blacksquare \end{align*}
(biii)
As n→∞,{ n \to \infty, } 12n+3→0{\displaystyle \frac{1}{2 n + 3} \to 0} so
S∞=13S_\infty = \frac{1}{3}
∣S∞−Sn∣<10−3∣13−(13−12n+3)∣<10−3∣12n+3∣<11000\begin{gather*} \left| S_\infty - S_n \right | < 10^{-3} \\ \left| \frac{1}{3} - \left( \frac{1}{3} - \frac{1}{2 n + 3} \right) \right| < 10^{-3} \\ \left| \frac{1}{2n+3} \right | < \frac{1}{1000} \\ \end{gather*}
Since 2n+3{2n+3} is positive,
2n+3>1000n>498.5\begin{align*} 2n+3 &> 1000 \\ n &> 498.5 \end{align*}
Smallest value of n=499  ■{n = 499 \;\blacksquare}