Find the sum of the series 1 cdot 2 cdot 3 + | vedclass

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(C) The $n$-th term of the series is $t_n = n(n + 1)(n + 2)$.We can write $t_n = n(n + 1)(n + 2) = \frac{1}{4} [n(n + 1)(n + 2)(n + 3) - (n - 1)n(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 $t_n = n(n + 1)(n + 2)$.We can write $t_n = n(n + 1)(n + 2) = \frac{1}{4} [n(n + 1)(n + 2)(n + 3) - (n - 1)n(n + 1)(n + 2)]$.Summing from $n = 1$ to $n$,we get a telescoping series:$S_n = \sum_{k=1}^{n} t_k = \frac{1}{4} [1 \cdot 2 \cdot 3 \cdot 4 - 0 + 2 \cdot 3 \cdot 4 \cdot 5 - 1 \cdot 2 \cdot 3 \cdot 4 + \dots + n(n + 1)(n + 2)(n + 3) - (n - 1)n(n + 1)(n + 2)]$.All intermediate terms cancel out,leaving:$S_n = \frac{1}{4} n(n + 1)(n + 2)(n + 3)$.

Find the sum of the series $1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + 3 \cdot 4 \cdot 5 + \dots$ up to $n$ terms.

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