If S_n denotes the sum of the products of the | vedclass

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(A) We know that $\sum_{1 \le i < j \le n} ij = \frac{1}{2} [(\sum_{i=1}^n i)^2 - \sum_{i=1}^n i^2]$.$S_n = \frac{1}{2} [(\frac{n(n+1)}{2})^2 - \frac{

Vedclass · Accepted Answer

$\frac{11e}{24}$

Vedclass · Answer

(A) We know that $\sum_{1 \le i $S_n = \frac{1}{2} [(\frac{n(n+1)}{2})^2 - \frac{n(n+1)(2n+1)}{6}]$$S_n = \frac{n(n+1)}{2} [\frac{n(n+1)}{4} - \frac{2n+1}{6}] = \frac{n(n+1)}{2} [\frac{3n^2+3n-4n-2}{12}] = \frac{n(n+1)(3n^2-n-2)}{24} = \frac{n(n+1)(n-1)(3n+2)}{24}$.For $n=0, 1$,$S_n = 0$. For $n \ge 2$,$\frac{S_n}{(n+1)!} = \frac{n(n+1)(n-1)(3n+2)}{24(n+1)n(n-1)(n-2)!} = \frac{3n+2}{24(n-2)!}$.$\sum_{n=2}^\infty \frac{3n+2}{24(n-2)!} = \frac{1}{24} \sum_{n=2}^\infty \frac{3(n-2)+8}{(n-2)!} = \frac{1}{24} [3 \sum_{n=2}^\infty \frac{1}{(n-3)!} + 8 \sum_{n=2}^\infty \frac{1}{(n-2)!}]$.Note: $\frac{1}{(n-3)!} = 0$ for $n=2$. So,$\frac{1}{24} [3 \sum_{n=3}^\infty \frac{1}{(n-3)!} + 8 \sum_{n=2}^\infty \frac{1}{(n-2)!}] = \frac{1}{24} [3e + 8e] = \frac{11e}{24}$.

If $S_n$ denotes the sum of the products of the first $n$ natural numbers taken two at a time,then $\sum\limits_{n = 0}^\infty \frac{S_n}{(n + 1)!} = $

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