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

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

Vedclass · Answer

(A) The $n^{th}$ term of the series is given by $a_n = n(n+1) = n^2 + n$. 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 $\sum_{k=1}^n k = \frac{n(n+1)}{2}$,we get: $S_n = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}$ Factor out $\frac{n(n+1)}{2}$: $S_n = \frac{n(n+1)}{2} \left( \frac{2n+1}{3} + 1 \right)$ $S_n = \frac{n(n+1)}{2} \left( \frac{2n+1+3}{3} \right)$ $S_n = \frac{n(n+1)}{2} \left( \frac{2n+4}{3} \right) = \frac{n(n+1) \cdot 2(n+2)}{2 \cdot 3}$ $S_n = \frac{n(n+1)(n+2)}{3}$

Find the sum to $n$ terms of the series $1 \times 2 + 2 \times 3 + 3 \times 4 + 4 \times 5 + \ldots$

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