2021 H2 Mathematics Paper 1 Question 5

Sigma Notation

Answers

(a)

−1x+2+3x+3−2x+4{- \frac{1}{x + 2} + \frac{3}{x + 3} - \frac{2}{x + 4}}

(b)

16+1n+3−2n+4{\frac{1}{6} + \frac{1}{n + 3} - \frac{2}{n + 4}}

(c)

16{\frac{1}{6}}

Full solutions

(a)

By the cover up rule,
x(x+2)(x+3)(x+4)=−1x+2+3x+3−2x+4  ■\begin{align*} & \frac{x}{(x + 2)(x + 3)(x + 4)} \\ & = - \frac{1}{x + 2} + \frac{3}{x + 3} - \frac{2}{x + 4} \; \blacksquare \end{align*}

(b)

∑r=1nr(r+2)(r+3)(r+4)=∑r=1n(−1r+2+3r+3−2r+4)=(−13+34−25−14+35−13−15+12−27−⋯−1n+3n+1−2n+2−1n+1+3n+2−2n+3−1n+2+3n+3−2n+4)=16+1n+3−2n+4  ■\begin{align*} & \sum_{r=1}^n \frac{r}{(r+2)(r+3)(r+4)} \\ &= \sum_{r=1}^n \left( -\frac{1}{r+2} + \frac{3}{r+3} - \frac{2}{r+4} \right) \\ & = \left(\def\arraystretch{1.5} \begin{array}{lclclc} - & \frac{1}{3} &+& \frac{3}{4} &-& \cancel{\frac{2}{5}} \\ - & \frac{1}{4} &+& \cancel{\frac{3}{5}} &-& \cancel{\frac{1}{3}} \\ - & \cancel{\frac{1}{5}} &+& \cancel{\frac{1}{2}} &-& \cancel{\frac{2}{7}} \\ - &&& \cdots && \\ - & \cancel{\frac{1}{n}} &+& \cancel{\frac{3}{n + 1}} &-& \cancel{\frac{2}{n + 2}} \\ - & \cancel{\frac{1}{n + 1}} &+& \cancel{\frac{3}{n + 2}} &-& \frac{2}{n + 3} \\ - & \cancel{\frac{1}{n + 2}} &+& \frac{3}{n + 3} &-& \frac{2}{n + 4} \end{array}\right) \\ &= \frac{1}{6} + \frac{1}{n + 3} - \frac{2}{n + 4} \; \blacksquare \end{align*}

(c)

As n→∞,{ n \to \infty, } 1n+3,2n+4→0{\displaystyle \frac{1}{n + 3},\frac{2}{n + 4} \to 0} so
∑r=1∞r(r+2)(r+3)(r+4)=lim⁡n→∞(16+1n+3−2n+4)=16  ■\begin{align*} & \sum_{r=1}^\infty \frac{r}{(r+2)(r+3)(r+4)} \\ & = \lim_{n \to \infty} \left( \frac{1}{6} + \frac{1}{n + 3} - \frac{2}{n + 4} \right) \\ & = \frac{1}{6} \; \blacksquare \end{align*}