Mathematics - Sequence & Series Question with Solution | TestHub

Given sum is $$\begin{array}{rl}& \frac{3}{12}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+\dots ..\\ & n\text{th term }={\mathrm{T}}_n\\ & =\frac{2n+1}{\frac{n(n+1)(2n+1)}{6}}=\frac{6}{n(n+1)}\\ & \text{ or }{\mathrm{T}}_n=6\left[\frac{1}{n}-\frac{1}{n+1}\right]\\ & \therefore {\mathrm{S}}_n=\\ & \sum {\mathrm{T}}_n=6\sum \frac{1}{n}-6\sum \frac{1}{n+1}=\frac{6n}{n}-\frac{6}{n+1}\\ & =6-\frac{6}{n+1}=\frac{6n}{n+1}\\ & S_{11}=\frac{6\times 11}{11+1}=\frac{66}{12}=\frac{33}{6}=\frac{11}{2}\\ & \end{array}$$