Sum of the n terms of the series frac{3}{1^2} | 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 of the first $n$ na

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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 of the first $n$ natural numbers,$\sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6}$.Substituting this into the expression for $T_n$:$T_n = \frac{2n + 1}{\frac{n(n + 1)(2n + 1)}{6}} = \frac{6}{n(n + 1)}$.Using partial fractions,we can write $T_n = 6 \left( \frac{1}{n} - \frac{1}{n + 1} \right)$.The sum of $n$ terms $S_n = \sum_{k=1}^{n} T_k = 6 \sum_{k=1}^{n} \left( \frac{1}{k} - \frac{1}{k + 1} \right)$.This is a telescoping series:$S_n = 6 \left[ (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + ... + (\frac{1}{n} - \frac{1}{n + 1}) \right]$.$S_n = 6 \left( 1 - \frac{1}{n + 1} \right) = 6 \left( \frac{n + 1 - 1}{n + 1} \right) = \frac{6n}{n + 1}$.

Sum of the $n$ terms of the series $\frac{3}{1^2} + \frac{5}{1^2 + 2^2} + \frac{7}{1^2 + 2^2 + 3^2} + ...$ is

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