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 $T_n = n(n + 1) = n^2 + n$.To find the sum of $n$ terms,we calculate $S_n = \sum_{k=1}^{n} T_k = \sum_

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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 $T_n = n(n + 1) = n^2 + n$.To find the sum of $n$ terms,we calculate $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (k^2 + k)$.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:$S_n = \frac{n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2}$.Factoring out $\frac{n(n + 1)}{6}$,we get:$S_n = \frac{n(n + 1)}{6} [(2n + 1) + 3] = \frac{n(n + 1)(2n + 4)}{6}$.$S_n = \frac{n(n + 1) \cdot 2(n + 2)}{6} = \frac{1}{3}n(n + 1)(n + 2)$.

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

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