The sum of 10 terms of the series frac{3}{1^{ | vedclass

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(B) The $n^{th}$ term of the series is given by $T_n = \frac{(n+1)^2 - n^2}{n^2(n+1)^2} = \frac{1}{n^2} - \frac{1}{(n+1)^2}$.For $n=1, 2, \ldots, 10$,

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$\frac{120}{121}$

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(B) The $n^{th}$ term of the series is given by $T_n = \frac{(n+1)^2 - n^2}{n^2(n+1)^2} = \frac{1}{n^2} - \frac{1}{(n+1)^2}$.For $n=1, 2, \ldots, 10$,the sum $S_{10}$ is:$S_{10} = \sum_{n=1}^{10} \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} \right)$.This is a telescoping series:$S_{10} = \left( \frac{1}{1^2} - \frac{1}{2^2} \right) + \left( \frac{1}{2^2} - \frac{1}{3^2} \right) + \ldots + \left( \frac{1}{10^2} - \frac{1}{11^2} \right)$.All intermediate terms cancel out:$S_{10} = 1 - \frac{1}{11^2} = 1 - \frac{1}{121} = \frac{120}{121}$.

The sum of $10$ terms of the series $\frac{3}{1^{2} \times 2^{2}}+\frac{5}{2^{2} \times 3^{2}}+\frac{7}{3^{2} \times 4^{2}}+\ldots$ is :

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