For all positive integral values of n,the val | vedclass

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(A) Let $T_k$ be the $k$-th term of the series,then $T_k = 3k(k + 1) = 3k^2 + 3k$.If $S_n$ denotes the sum of the first $n$ terms,then $S_n = \sum_{k=

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$n(n + 1)(n + 2)$

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(A) Let $T_k$ be the $k$-th term of the series,then $T_k = 3k(k + 1) = 3k^2 + 3k$.If $S_n$ denotes the sum of the first $n$ terms,then $S_n = \sum_{k=1}^n T_k = \sum_{k=1}^n (3k^2 + 3k)$.$S_n = 3 \sum_{k=1}^n k^2 + 3 \sum_{k=1}^n k$.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:$S_n = 3 \left[ \frac{n(n+1)(2n+1)}{6} \right] + 3 \left[ \frac{n(n+1)}{2} \right]$.$S_n = \frac{n(n+1)(2n+1)}{2} + \frac{3n(n+1)}{2} = \frac{n(n+1)}{2} [ (2n+1) + 3 ]$.$S_n = \frac{n(n+1)}{2} [ 2n + 4 ] = \frac{n(n+1)}{2} \cdot 2(n + 2) = n(n+1)(n+2)$.

For all positive integral values of $n$,the value of $3 \cdot 1 \cdot 2 + 3 \cdot 2 \cdot 3 + 3 \cdot 3 \cdot 4 + \dots + 3 \cdot n \cdot (n + 1)$ is

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