2019 H2 Mathematics Paper 1 Question 6

Sigma Notation

Answers

(i)

12(2r−1)−12(2r+1){\frac{1}{2 \left( 2 r - 1 \right)}-\frac{1}{2 \left( 2 r + 1 \right)}}
12−12(2n+1){\frac{1}{2} - \frac{1}{2 \left( 2 n + 1 \right)}}

(ii)

142{\frac{1}{42}}

Full solutions

(i)

By the cover up rule,
14r2−1=1(2r+1)(2r−1)=12(2r−1)−12(2r+1)  ■\begin{align*} &\frac{1}{4 r^2 - 1} \\ &= \frac{1}{(2 r + 1)(2 r - 1)} \\ &= \frac{1}{2 \left( 2 r - 1 \right)} - \frac{1}{2 \left( 2 r + 1 \right)} \; \blacksquare \end{align*}
∑r=1n24r2−1=∑r=1n(12(2r−1)−12(2r+1))=12−16+16−110+110−114+⋯+12(2n−5)−12(2n−3)+12(2n−3)−12(2n−1)+12(2n−1)−12(2n+1)=12−12(2n+1)  ■\begin{align*} & \sum_{r=1}^n \frac{2}{4 r^2 - 1} \\ &= \sum_{r=1}^n \left( \frac{1}{2 \left( 2 r - 1 \right)} - \frac{1}{2 \left( 2 r + 1 \right)}\right) \\ & = \def\arraystretch{1.5} \begin{array}{lclc} & \frac{1}{2} &-& \cancel{\frac{1}{6}} \\ + & \cancel{\frac{1}{6}} &-& \cancel{\frac{1}{10}} \\ + & \cancel{\frac{1}{10}} &-& \cancel{\frac{1}{14}} \\ + & & \cdots & \\ + & \cancel{\frac{1}{2 \left( 2 n - 5 \right)}} &-& \cancel{\frac{1}{2 \left( 2 n - 3 \right)}} \\ + & \cancel{\frac{1}{2 \left( 2 n - 3 \right)}} &-& \cancel{\frac{1}{2 \left( 2 n - 1 \right)}} \\ + & \cancel{\frac{1}{2 \left( 2 n - 1 \right)}} &-& \frac{1}{2 \left( 2 n + 1 \right)} \end{array} \\ \\ &= \frac{1}{2} - \frac{1}{2 \left( 2 n + 1 \right)} \; \blacksquare \end{align*}

(ii)

As n→∞,{ n \to \infty, } 12(2n+1)→0{\displaystyle \frac{1}{2 \left( 2 n + 1 \right)} \to 0} so
∑r=1∞14r2−1=12\sum_{r=1}^\infty \frac{1}{4r^2-1}= \frac{1}{2}
∑r=11∞14r2−1=∑r=1∞14r2−1−∑r=11014r2−1=12−(12−12(2(10)+1))=142  ■\begin{align*} & \sum_{r=11}^\infty \frac{1}{4 r^2 - 1} \\ & = \sum_{r=1}^\infty \frac{1}{4 r^2 - 1} - \sum_{r=1}^{10} \frac{1}{4 r^2 - 1} \\ & = \frac{1}{2} - \left( \frac{1}{2} - \frac{1}{2\left(2(10)+1\right)} \right) \\ &= \frac{1}{42} \; \blacksquare \end{align*}