The sum $\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+\ldots$. upto 11-terms is:

Question

Aicharya · Accepted Answer

Given sum is $ \begin{aligned} & \frac{3}{12}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+\ldots . . \ & n 	ext {th term }=\mathrm{T}_n \ & =\frac{2 n+1}{\frac{n(n+1)(2 n+1)}{6}}=\frac{6}{n(n+1)} \ & 	ext { or } \mathrm{T}_n=6\left[\frac{1}{n}-\frac{1}{n+1}ight] \ & 	herefore \quad \mathrm{S}_n= \ & \sum \mathrm{T}_n=6 \sum \frac{1}{n}-6 \sum \frac{1}{n+1}=\frac{6 n}{n}-\frac{6}{n+1} \ & =6-\frac{6}{n+1}=\frac{6 n}{n+1} \ & S_{11}=\frac{6 	imes 11}{11+1}=\frac{66}{12}=\frac{33}{6}=\frac{11}{2} \ & \end{aligned} $

The sum $\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+\ldots$. upto 11-terms is:

Solution