Series Demystified 3

$S=\large \sum_{n=1}^{48} \left [ \frac {n!(n^2+n-1)}{n\sqrt{n+1}+\sqrt n}\right]$

$= \large \sum_{n=1}^{48} \left [ \frac {n!(n^2+n-1)(n\sqrt{n+1}-\sqrt n)}{(n\sqrt{n+1}+\sqrt n)(n\sqrt{n+1}-\sqrt n)}\right]$

$= \large \sum_{n=1}^{48} \left [ \frac {n!(n^2+n-1)(n\sqrt{n+1}-\sqrt n)}{n(n^2+n-1)}\right]$

$= \large \sum_{n=1}^{48} \left [ n!\sqrt{n+1}-(n-1)!\sqrt{n}\right]$

$S=48!\sqrt{49}-0!\sqrt{1}=48!\times 7 -1$

$\implies \frac{S+1}{48!}=\boxed{7}$

Let us demystify the expression that is going to be summated.

$\dfrac {n!(n^2 + n - 1)}{n\sqrt{n+1} + \sqrt{n}}$

$\Rightarrow \dfrac{n!(n^2 + n - 1)}{\sqrt{n}(\sqrt{n^2 + n}+ 1)}$

$\Rightarrow \dfrac{n!(\sqrt{n^2 + n} - 1)}{\sqrt{n}}$

$\Rightarrow n!(\sqrt{n+1} - \dfrac{1}{\sqrt{n}})$

$\Rightarrow \dfrac{(n+1)!}{\sqrt{n+1}} - \dfrac{n!}{\sqrt{n}}$

We have it yet again, a telescopic series

$S=\displaystyle \sum_{n=1}^{48} \dfrac{(n+1)!}{\sqrt{n+1}} - \dfrac{n!}{\sqrt{n}}$

$S=\dfrac{49!}{\sqrt{49}} - \dfrac{1!}{\sqrt{1}}$

Required answer is:

$\dfrac{S+1}{48!} = \dfrac{\frac{49!}{7}}{48!} = 7$