The sum of n terms of the series

Sum of $n$ term of $\frac{3}{1^{2}}+\frac{5}{1^{2}+2^{2}}+\frac{7}{1^{2}+2^{2}+3^{2}}$

here nth term tn for above series is

i. e $T_{n}=\frac{2 n+1}{1^{2}+2^{2}+\ldots .+n^{2}}$

$T_{n}=\frac{2 n+1}{\frac{\{n(n+1)(2 n+1)\}}{6}}=\frac{2 n+1}{\frac{\{(n+1)(2 n+1)\}}{6}}$

$T_{n}=\frac{6}{n(n+1)}$

i. e $T_{r}=\frac{6}{r(r+1)}$

Hence $\sum_{r=1}^{n} T_{r}=\sum_{r=1}^{n} \frac{6}{r(r+1)}$

$=6 \sum_{r=1}^{n}\left[\frac{1}{r(r+1)}\right]$

$=\sum_{r=1}^{n} 6\left[\frac{1}{r}-\frac{1}{r+1}\right]$

$=6\left[\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{n}-\frac{1}{n+1}\right)\right]$

$=6\left[1-\frac{1}{n+1}\right]=6\left[\frac{n}{n+1}\right]$

Hence, $S_{n}$ (sum of $n$ term) $=\frac{6 n}{n+1}$