I need help in solving this question. If $n \in N$,$n\ge 3$ then the value of $(1)n-\frac{(n-1)}{1!}+\frac{(n-1)(n-2)^2}{2!}-\frac{(n-1)(n-2)(n-3)^2}{3!}+.....$ upto n terms
any help hint will do.
My attempt: $S=\sum_{r=0}^n(n-r) {(n-1)\choose r}$. after this i am not able to do it.
Let $ n\in\mathbb{N}^{*} : $
Denoting $ S_{n} $ your summation, $ R_{n}=\sum\limits_{r=0}^{n-1}{\left(n-1-r\right)\binom{n-1}{r}} $, we have :
$$ S_{n}=\sum_{r=0}^{n-1}{\left(n-r\right)\binom{n-1}{r}}=\sum_{r=0}^{n-1}{\binom{n-1}{r}}+\sum_{r=0}^{n-1}{\left(n-1-r\right)\binom{n-1}{r}}=2^{n-1}+R_{n} $$
Now making a change of index by symmetry in $ R_{n} $, we get that : $$ R_{n}=\sum_{r=0}^{n-1}{\left(n-1-r\right)\binom{n-1}{r}}=\sum_{r=0}^{n-1}{r\binom{n-1}{r}} $$
Thus, $$ 2R_{n}=\sum_{r=0}^{n-1}{\left(n-1\right)\binom{n-1}{r}}=\left(n-1\right)2^{n-1} $$
Hence $$ S_{n}=2^{n-1}+\left(n-1\right)2^{n-2}=\left(n+1\right)2^{n-2} $$