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 the sum of the first $n$ natural numbers:$T_n = 1 + 2 + 3 + \dots + n = \frac{n(n + 1)}{2}$The sum of

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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 the sum of the first $n$ natural numbers:$T_n = 1 + 2 + 3 + \dots + n = \frac{n(n + 1)}{2}$The sum of $n$ terms $S_n$ is the sum of $T_n$ from $k=1$ to $n$:$S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \frac{k(k + 1)}{2} = \frac{1}{2} \left[ \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k \right]$Using the standard summation formulas $\sum k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum k = \frac{n(n+1)}{2}$:$S_n = \frac{1}{2} \left[ \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \right]$Factor out $\frac{n(n+1)}{2}$:$S_n = \frac{1}{2} \cdot \frac{n(n+1)}{2} \left[ \frac{2n+1}{3} + 1 \right] = \frac{n(n+1)}{4} \left[ \frac{2n+1+3}{3} \right]$$S_n = \frac{n(n+1)}{4} \cdot \frac{2n+4}{3} = \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 will be:

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