The sum of the series 1 times 2 times 3 + 2 t | vedclass

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(C) The $n^{th}$ term of the series is $T_n = n(n + 1)(n + 2)$.Expanding this,we get $T_n = n(n^2 + 3n + 2) = n^3 + 3n^2 + 2n$.To find the sum $S_n$,w

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$\frac{1}{4}n(n + 1)(n + 2)(n + 3)$

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(C) The $n^{th}$ term of the series is $T_n = n(n + 1)(n + 2)$.Expanding this,we get $T_n = n(n^2 + 3n + 2) = n^3 + 3n^2 + 2n$.To find the sum $S_n$,we calculate $\sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (k^3 + 3k^2 + 2k)$.Using standard summation formulas:$S_n = \left[ \frac{n(n + 1)}{2} \right]^2 + 3 \left[ \frac{n(n + 1)(2n + 1)}{6} \right] + 2 \left[ \frac{n(n + 1)}{2} \right]$.Simplifying the expression:$S_n = \frac{n^2(n + 1)^2}{4} + \frac{n(n + 1)(2n + 1)}{2} + n(n + 1)$.Factoring out $\frac{n(n + 1)}{4}$:$S_n = \frac{n(n + 1)}{4} [n(n + 1) + 2(2n + 1) + 4]$.$S_n = \frac{n(n + 1)}{4} [n^2 + n + 4n + 2 + 4] = \frac{n(n + 1)}{4} [n^2 + 5n + 6]$.$S_n = \frac{n(n + 1)(n + 2)(n + 3)}{4}$.

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

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