Question

The sum of n terms of the series $\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+$_________ is _______________.

Answer 1

Sum of n term of $\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}$
here n^th term t_n for above series is
$\begin{array}{rcl}\mathrm{i}.\mathrm{e}{T}_n& =& \frac{2n+1}{1^2+2^2+....+n^2}\\ T_n& =& \frac{2n+1}{{\displaystyle \frac{\left\{n\left(n+1\right)\left(2n+1\right)\right\}}{6}}}=\frac{2n+1}{{\displaystyle \frac{\left\{\left(n+1\right)\left(2n+1\right)\right\}}{6}}}\\ T_n& =& \frac{6}{n\left(n+1\right)}\\ \mathrm{i}.\mathrm{e}{T}_r& =& \frac{6}{r\left(r+1\right)}\end{array}$
$\begin{array}{rcl}\mathrm{Hence}\sum _{r=1}^n{T}_r& =& \sum _{r=1}^n\frac{6}{r\left(r+1\right)}\\ & =& 6\sum _{r=1}^n\left[\frac{1}{r\left(r+1\right)}\right]\\ & =& \sum _{r=1}^{n}6\left[\frac{1}{r}-\frac{1}{r+1}\right]\\ & =& 6\left[\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+....+\left(\frac{1}{n}-\frac{1}{n+1}\right)\right]\\ & =& 6\left[1-\frac{1}{n+1}\right]=6\left[\frac{n}{n+1}\right]\end{array}$
$\mathrm{Hence},S_{\mathit{n}\mathit{}}\left(\mathrm{sum}\mathrm{of}n\mathrm{term}\right)=\frac{6n}{n+1}$