The value of the sum 1 times 2 times 3 + 2 ti | vedclass

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(D) The $n$-th term of the series is $T_n = n(n+1)(n+2)$.Expanding this,we get $T_n = n(n^2 + 3n + 2) = n^3 + 3n^2 + 2n$.The sum of $n$ terms is $S_n

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

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(D) The $n$-th term of the series is $T_n = n(n+1)(n+2)$.Expanding this,we get $T_n = n(n^2 + 3n + 2) = n^3 + 3n^2 + 2n$.The sum of $n$ terms is $S_n = \sum_{k=1}^n T_k = \sum_{k=1}^n (k^3 + 3k^2 + 2k)$.Using standard summation formulas:$\sum_{k=1}^n k^3 = \frac{n^2(n+1)^2}{4}$,$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$,and $\sum_{k=1}^n k = \frac{n(n+1)}{2}$.Substituting these:$S_n = \frac{n^2(n+1)^2}{4} + 3 \cdot \frac{n(n+1)(2n+1)}{6} + 2 \cdot \frac{n(n+1)}{2}$$S_n = \frac{n(n+1)}{2} \left[ \frac{n(n+1)}{2} + (2n+1) + 2 \right]$$S_n = \frac{n(n+1)}{2} \left[ \frac{n^2+n+4n+2+4}{2} \right] = \frac{n(n+1)(n^2+5n+6)}{4}$Since $n^2+5n+6 = (n+2)(n+3)$,we have $S_n = \frac{n(n+1)(n+2)(n+3)}{4}$.

The value of the sum $1 \times 2 \times 3 + 2 \times 3 \times 4 + 3 \times 4 \times 5 + \ldots$ up to $n$ terms is equal to

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