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Q.
The sum $\frac{3}{1^{2}}+\frac{5}{1^{2}+2^{2}}+\frac{7}{1^{2}+2^{2}+3^{2}}+....$ upto 11 terms is:
Sequences and Series
Solution:
Given sum is
$\frac{3}{1^{2}}+\frac{5}{1^{2}+2^{2}}+\frac{7}{1^{2}+2^{2}+3^{2}}+....$
$nth$ term $=T_{n}=\frac{2n+1}{\frac{n\left(n+1\right)\left(2n+1\right)}{6}}=\frac{6}{n\left(n+1\right)}$
or $T_{n}=6\left[\frac{1}{n}-\frac{1}{n+1}\right]$
$\therefore S_{n}=\sum\,T_{n}=6\,\sum \frac{1}{n}-6\,\sum \frac{1}{n+1}=\frac{6n}{n}-\frac{6}{n+1}$
$=6-\frac{6}{n+1}=\frac{6n}{n+1}$
So, sum upto 11 terms means
$S_{11}=\frac{6\times11}{11+1}=\frac{66}{12}=\frac{33}{6}=\frac{11}{2}$