# find sum 1/6 + 1/24 + 1/60.....1/r^3 – r. How do we answer this question. Please answer ASAP

Grade:11

## 1 Answers

jagdish singh singh
173 Points
6 years ago
$\hspace{-0.9 cm.} Let t_{r} = \frac{1}{r(r^2-1)} = \frac{1}{(r-1)r(r+1)} = \frac{1}{2}\left[\frac{1}{(r-1)r}-\frac{1}{r(r+1)}\right]\\\\So \sum^{n}_{r=2}t_{r} =\frac{1}{2}\sum^{n}_{r=2}\left[\frac{1}{(r-1)r}-\frac{1}{r(r+1)}\right] \\\\Now Using Telescoping Sum, We get \sum^{n}_{r=2}t_{r}=\frac{1}{2}\left[1-\frac{1}{n(n+1)}\right]\\\\So we get \sum^{n}_{r=1}t_{r} = \frac{n^2+n+1}{2n(n+1)} and \lim_{n\rightarrow 0}\sum^{n}_{r=1}t_{r} = \frac{1}{2}.$