Question

If $S_n$ represents the sum of the product of the first n natural numbers taken two at a time, then $\frac{2}{3!}+\frac{11}{4!}+...+\frac{S_{n-1}}{n!}+...=\frac{a}{24}e$ Find a

Answer 1

Since $S_n=1.2+1.3+....+1.n+2.3+2.4+2.n+3.4+3.5+...+3.n+...+(n-1).n$

But $(1+2+3+...+n{)}^2=1^2+2^2+3^2+...+n^2+2S_n$

$\to$ $(\sum n{)}^2=\sum n^2+2S_n$

$\therefore S_n=\frac{1}{2}[{(\sum n)}^2+\sum n^2]$

$=\frac{1}{2}[{\left(\frac{n(n+1)}{2}\right)}^2-\frac{n(n+1)(2n+1)}{6}]$

$=\frac{1}{2}[\left(\frac{n^2(n+1{)}^2}{4}\right)-\frac{n(n+1)(2n+1)}{6}]$

$=\frac{n(n+1)}{24}\left[3n\right(n+1)-2(2n+1\left)\right]$

$=\frac{n(n+1)}{24}(3n^2-n-2)$

$=\frac{n(n+1)(n-1)(3n+2)}{24}$

$\therefore T_{n+1}=\frac{S_n}{(n+1)!}$

$=\frac{(n-1)n(n+1)(3n+2)}{24(n+1)!}$

$=\frac{(n-1)n(n+1)(3n+2)}{24(n+1)n(n-1)(n-2)!}$

$=\frac{(3n+2)}{24(n-2)!}$

$=\frac{3(n-2)+8}{24(n-1)!}$

$=\frac{3(n-2)}{24(n-2)!}+\frac{8}{24(n-2)!}$

$=\frac{3(n-2)}{24(n-2)(n-3)!}+\frac{8}{24(n-2)!}$

$=\frac{3}{24(n-3)!}+\frac{8}{24(n-2)!}$

Putting n=1, 2, 3, 4,. We get the series

$=\frac{3}{24}(\frac{1}{(-3)!}+\frac{1}{(-2)!}+\frac{1}{(-1)!}+\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+...)+\frac{8}{24}(\frac{1}{(-2)!}+\frac{1}{(-1)!}+\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+)$

$=\frac{3}{24}(0+0+0+e)+\frac{8}{24}(0+0+e)$

$=\frac{3e}{24}+\frac{8e}{24}$

$=\frac{11e}{24}$

Ans: 11