Let S_{n} = 1 cdot (n-1) + 2 cdot (n-2) + 3 c | vedclass

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(A) The general term of the sum $S_{n}$ is $T_{r} = r(n-r)$ for $r = 1$ to $n-1$.$S_{n} = \sum_{r=1}^{n-1} (nr - r^{2}) = n \sum_{r=1}^{n-1} r - \sum_

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$\frac{e-1}{3}$

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(A) The general term of the sum $S_{n}$ is $T_{r} = r(n-r)$ for $r = 1$ to $n-1$.$S_{n} = \sum_{r=1}^{n-1} (nr - r^{2}) = n \sum_{r=1}^{n-1} r - \sum_{r=1}^{n-1} r^{2}$.$S_{n} = n \frac{(n-1)n}{2} - \frac{(n-1)n(2n-1)}{6} = \frac{n(n-1)}{6} [3n - (2n-1)] = \frac{n(n-1)(n+1)}{6}$.Now,consider the term inside the summation: $\frac{2 S_{n}}{n!} - \frac{1}{(n-2)!} = \frac{2 n(n-1)(n+1)}{6 n(n-1)(n-2)!} - \frac{1}{(n-2)!}$.$= \frac{n+1}{3(n-2)!} - \frac{1}{(n-2)!} = \frac{n+1-3}{3(n-2)!} = \frac{n-2}{3(n-2)!} = \frac{1}{3(n-3)!}$.Summing from $n=4$ to $\infty$: $\sum_{n=4}^{\infty} \frac{1}{3(n-3)!} = \frac{1}{3} \sum_{k=1}^{\infty} \frac{1}{k!} = \frac{1}{3} (e-1)$.

Let $S_{n} = 1 \cdot (n-1) + 2 \cdot (n-2) + 3 \cdot (n-3) + \dots + (n-1) \cdot 1$,for $n \geq 4$. The sum $\sum_{n=4}^{\infty} \left( \frac{2 S_{n}}{n!} - \frac{1}{(n-2)!} \right)$ is equal to:

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