The sum of n terms of the series whose n^{th} | vedclass

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(A) Given the $n^{th}$ term is $T_n = n(n + 1) = n^2 + n$.The sum of $n$ terms is $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (k^2 + k)$.Using the stan

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$\frac{n(n + 1)(n + 2)}{3}$

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(A) Given the $n^{th}$ term is $T_n = n(n + 1) = n^2 + n$.The sum of $n$ terms is $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (k^2 + k)$.Using the standard summation formulas $\sum k^2 = \frac{n(n + 1)(2n + 1)}{6}$ and $\sum k = \frac{n(n + 1)}{2}$,we get:$S_n = \frac{n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2}$.Factor out $\frac{n(n + 1)}{2}$:$S_n = \frac{n(n + 1)}{2} \left( \frac{2n + 1}{3} + 1 \right)$.$S_n = \frac{n(n + 1)}{2} \left( \frac{2n + 1 + 3}{3} \right) = \frac{n(n + 1)(2n + 4)}{6}$.$S_n = \frac{n(n + 1) \cdot 2(n + 2)}{6} = \frac{n(n + 1)(n + 2)}{3}$.

The sum of $n$ terms of the series whose $n^{th}$ term is $n(n + 1)$ is equal to

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