Question

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

• A. $\frac{7}{2}$
• B. $\frac{11}{4}$
• C. $\frac{11}{2}$
• D. $\frac{60}{11}$

Answer 1

The correct option is C $\frac{11}{2}$

$\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+.....\text{upto 11 terms}$

General term $=\frac{2n+1}{\left(\frac{n(n+1)(2n+1)}{6}\right)}=\frac{6}{n(n+1)}$

So the series will be-

$\frac{6}{1\cdot 2}+\frac{6}{2\cdot 3}+\frac{6}{3\cdot 4}+.....+\frac{6}{11\cdot 12}$

$=6(\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+.....+\frac{1}{11\cdot 12})$

$=6[(\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+.....+(\frac{1}{11}-\frac{1}{12})]$

$=6(1-\frac{1}{12})$

$=6\times \frac{11}{12}$

$=\frac{11}{2}$