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 $T_{n} = \frac{\sum_{k=1}^{n} k^{2}}{\sum_{k=1}^{n} k}$.Using the formulas $\sum k^{2} = \frac{n(n+1)(2n+

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$\frac{n(n+2)}{3}$

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(C) The $n$th term of the series is given by $T_{n} = \frac{\sum_{k=1}^{n} k^{2}}{\sum_{k=1}^{n} k}$.Using the formulas $\sum k^{2} = \frac{n(n+1)(2n+1)}{6}$ and $\sum k = \frac{n(n+1)}{2}$,we get:$T_{n} = \frac{n(n+1)(2n+1)}{6} 	imes \frac{2}{n(n+1)} = \frac{2n+1}{3}$.The sum of the first $n$ terms is $S_{n} = \sum_{k=1}^{n} T_{k} = \sum_{k=1}^{n} \frac{2k+1}{3}$.$S_{n} = \frac{1}{3} \left( 2 \sum_{k=1}^{n} k + \sum_{k=1}^{n} 1 ight)$.$S_{n} = \frac{1}{3} \left( 2 \cdot \frac{n(n+1)}{2} + n ight) = \frac{1}{3} (n^{2} + n + n) = \frac{n^{2}+2n}{3} = \frac{n(n+2)}{3}$.

The sum of the first $n$ terms of the series $\frac{1^{2}}{1}+\frac{1^{2}+2^{2}}{1+2}+\frac{1^{2}+2^{2}+3^{2}}{1+2+3}+\ldots$ is

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