• # question_answer The sum $\frac{3}{{{1}^{2}}}+\frac{5}{{{1}^{2}}+{{2}^{2}}}+\frac{7}{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}}+...$ upto 11 ?terms is:     JEE Main  Online Paper (Held On 22 April 2013) A)  $\frac{7}{2}$                                      B)  $\frac{11}{4}$ C)  $\frac{11}{2}$                                    D)  $\frac{60}{11}$

Given sum is $\frac{3}{12}+\frac{5}{{{1}^{2}}+{{2}^{2}}}+\frac{7}{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}}+.....$ $nth\,\,term={{T}_{n}}$ $=\frac{2n+1}{\frac{n(n+1)(2n+1)}{6}}=\frac{6}{n(n+1)}$ 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 ${{S}_{11}}=\frac{6\times 11}{11+1}=\frac{66}{12}=\frac{33}{6}=\frac{11}{2}$
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