The sum of the first n terms of the series fr | vedclass

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(C) The $n^{th}$ term of the series is given by $T_n = \frac{2n + 1}{\sum_{k=1}^{n} k^2}$.Using the formula for the sum of squares,$\sum_{k=1}^{n} k^2

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$\frac{6n}{n + 1}$

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(C) The $n^{th}$ term of the series is given by $T_n = \frac{2n + 1}{\sum_{k=1}^{n} k^2}$.Using the formula for the sum of squares,$\sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6}$.Thus,$T_n = \frac{2n + 1}{\frac{n(n + 1)(2n + 1)}{6}} = \frac{6}{n(n + 1)}$.Using partial fractions,$T_n = 6 \left[ \frac{1}{n} - \frac{1}{n + 1} \right]$.The sum of $n$ terms is $S_n = \sum_{k=1}^{n} T_k = 6 \sum_{k=1}^{n} \left( \frac{1}{k} - \frac{1}{k + 1} \right)$.This is a telescoping series: $S_n = 6 \left[ (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{n} - \frac{1}{n + 1}) \right]$.$S_n = 6 \left[ 1 - \frac{1}{n + 1} \right] = 6 \left[ \frac{n + 1 - 1}{n + 1} \right] = \frac{6n}{n + 1}$.

The sum of the first $n$ terms of the series $\frac{3}{1^2} + \frac{5}{1^2 + 2^2} + \frac{7}{1^2 + 2^2 + 3^2} + \dots$ is $.........$.

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