The sum of the series 1 + (1 + 2) + (1 + 2 + | vedclass

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(D) The $n^{th}$ term of the series is given by $T_n = 1 + 2 + 3 + \dots + n = \frac{n(n + 1)}{2}$.The sum of $n$ terms is $S_n = \sum_{k=1}^{n} T_k =

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$\frac{n(n + 1)(n + 2)}{6}$

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(D) The $n^{th}$ term of the series is given by $T_n = 1 + 2 + 3 + \dots + n = \frac{n(n + 1)}{2}$.The sum of $n$ terms is $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \frac{k(k + 1)}{2}$.$S_n = \frac{1}{2} \left( \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k \right)$.Using the standard summation formulas $\sum k^2 = \frac{n(n + 1)(2n + 1)}{6}$ and $\sum k = \frac{n(n + 1)}{2}$:$S_n = \frac{1}{2} \left( \frac{n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2} \right)$.$S_n = \frac{n(n + 1)}{4} \left( \frac{2n + 1}{3} + 1 \right) = \frac{n(n + 1)}{4} \left( \frac{2n + 4}{3} \right)$.$S_n = \frac{n(n + 1) \cdot 2(n + 2)}{12} = \frac{n(n + 1)(n + 2)}{6}$.

The sum of the series $1 + (1 + 2) + (1 + 2 + 3) + \dots$ up to $n$ terms is:

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