The sum of n terms of the following series 1 | vedclass

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(B) The $n^{th}$ term of the series is given by $T_n = n(n + 1) = n^2 + n$.The sum of $n$ terms is $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (k^2 + k

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

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(B) The $n^{th}$ term of the series is given by $T_n = n(n + 1) = n^2 + n$.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}$$S_n = \frac{n(n+1)}{2} \left[ \frac{2n+1}{3} + 1 ight]$$S_n = \frac{n(n+1)}{2} \left[ \frac{2n+1+3}{3} ight] = \frac{n(n+1)}{2} \left[ \frac{2n+4}{3} ight]$$S_n = \frac{n(n+1) 	imes 2(n+2)}{2 	imes 3} = \frac{1}{3}n(n + 1)(n + 2)$.

The sum of $n$ terms of the following series $1 \times 2 + 2 \times 3 + 3 \times 4 + 4 \times 5 + \dots$ is:

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