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

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

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

Vedclass · Answer

(D) The $k$-th term of the series is $T_k = 1 + 2 + 3 + \dots + k = \frac{k(k + 1)}{2}$.The sum of $n$ terms is $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \frac{k(k + 1)}{2}$.$S_n = \frac{1}{2} \sum_{k=1}^{n} (k^2 + k) = \frac{1}{2} [\sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k]$.Using the formulas $\sum k^2 = \frac{n(n + 1)(2n + 1)}{6}$ and $\sum k = \frac{n(n + 1)}{2}$:$S_n = \frac{1}{2} [\frac{n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2}] = \frac{n(n + 1)}{4} [\frac{2n + 1}{3} + 1]$.$S_n = \frac{n(n + 1)}{4} [\frac{2n + 4}{3}] = \frac{n(n + 1) \cdot 2(n + 2)}{12} = \frac{n(n + 1)(n + 2)}{6}$.

Find the sum of the series $1 + (1 + 2) + (1 + 2 + 3) + \dots$ up to $n$ terms.

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