The sum of the series 1 cdot 2 cdot 3 + 2 cdo | vedclass

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

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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 given by $T_n = n(n + 1)(n + 2)$.Expanding this,we get $T_n = n(n^2 + 3n + 2) = n^3 + 3n^2 + 2n$.The sum of $n$ terms is $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (k^3 + 3k^2 + 2k)$.Using the standard summation formulas:$\sum k^3 = \left[ \frac{n(n+1)}{2} \right]^2$,$\sum k^2 = \frac{n(n+1)(2n+1)}{6}$,and $\sum k = \frac{n(n+1)}{2}$.$S_n = \left[ \frac{n(n+1)}{2} \right]^2 + 3 \cdot \frac{n(n+1)(2n+1)}{6} + 2 \cdot \frac{n(n+1)}{2}$.$S_n = \frac{n^2(n+1)^2}{4} + \frac{n(n+1)(2n+1)}{2} + n(n+1)$.Taking $\frac{n(n+1)}{4}$ as a common factor:$S_n = \frac{n(n+1)}{4} [n(n+1) + 2(2n+1) + 4]$.$S_n = \frac{n(n+1)}{4} [n^2 + n + 4n + 2 + 4] = \frac{n(n+1)}{4} [n^2 + 5n + 6]$.Since $n^2 + 5n + 6 = (n+2)(n+3)$,we get $S_n = \frac{1}{4}n(n + 1)(n + 2)(n + 3)$.

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

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