If the n^{th} term of a series is n(n + 1),th | vedclass

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(A) The $n^{th}$ term is $a_n = n(n + 1) = n^2 + n$. The sum of $n$ terms is $S_n = \sum_{k=1}^{n} a_k = \sum_{k=1}^{n} (k^2 + k)$. Using the standard

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

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

If the $n^{th}$ term of a series is $n(n + 1)$,then the sum of its $n$ terms is......

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