Prove equation including sum of sequence and combinatorics

Question

Prove that:

$\sum\limits_{i=1}^n \frac{i^2*n!}{i!(n-i)!} = n*(n+1)*2^{n-2}$

I most probably have to use induction, but as much as I've tried, it doesn't bring me closer to a solution. I'm probably just over thinking it or can't see something in plain sight.

Hints are ideas for a method are welcome!

Answer 1

For $2\le r\le n,$

$$\frac{r^2\cdot n!}{r!\cdot(n-r)!}$$

$$=\frac{\{r(r-1)+r\}n!}{r!\cdot(n-r)!}$$

$$=\frac{r(r-1)\cdot n!}{r!\cdot(n-r)!}+\frac{r\cdot n!}{r!\cdot(n-r)!}$$

$$=n(n-1)\frac{r(r-1)\cdot (n-2)!}{r(r-1)\cdot(r-2)!\cdot\{(n-2)-(r-2)\}!}+n\frac{r\cdot (n-1)!}{r\cdot (r-1)!\{(n-1)-(r-1)\}!}$$

$$=n(n-1)\binom{n-2}{r-2}+n\binom{n-1}{r-1}$$

For $r=1,$

$$\frac{r^2\cdot n!}{r!\cdot(n-r)!}=\frac{n!}{(n-1)!}=n\binom{n-1}0$$

Can you take it from here?

Answer 2

$$\sum_{i=1}^n\frac{i^2n!}{i!(n-i)!}=\sum_{i=1}^ni^2\binom{n}{i}=\sum_{i=1}^ni^2\frac{n}{i}\binom{n-1}{i-1}=n\sum_{i=1}^ni\binom{n-1}{i-1}=$$ $$n\sum_{j=0}^{n-1}(j+1)\binom{n-1}{j}=n\sum_{j=0}^{n-1}j\binom{n-1}{j}+n\sum_{j=0}^{n-1}\binom{n-1}{j}=$$ $$=n\sum_{j=1}^{n-1}j\frac{n-1}{j}\binom{n-2}{j-1}+n2^{n-1}=n(n-1)\sum_{j=1}^{n-1}\binom{n-2}{j-1}+n2^{n-1}=$$ $$=n(n-1)\sum_{k=0}^{n-2}\binom{n-2}{k}+n2^{n-1}=n(n-1)2^{n-2}+2n2^{n-2}=$$ $$=n2^{n-2}(n-1+2)=n(n+1)2^{n-2}$$

Answer 3

Hint:

Note that $\frac{n!}{i!(n-i)!}=\binom{n}{i}$,

So $\sum_{i=1}^{n}\frac{i^2 n!}{i!(n-i)!}=\sum_{i=1}^{n}\binom{n}{i}i^2$

Consider $(1+x)^n=\sum_{i=0}^{n}\binom{n}{i}x^i$ (*)

Taking the derivative both of sides of (*),we have:

$n(1+x)^{n-1}=\sum_{i=1}^{n}\binom{n}{i}ix^{i-1}$ (**)

Mutiplying both of sides of (**) with $x$ we have:

$nx(1+x)^{n-1}=\sum_{i=1}^{n}\binom{n}{i}ix^{i}$

Taking the derivative both of sides of the above equality again gives us:

$n\left ((1+x)^{n-1}+x(n-1)(1+x)^{n-2} \right )=\sum_{i=1}^{n}\binom{n}{i}i^2x^{i}$

Now, let $x=1$ we have:

$\sum_{i=1}^{n}\frac{i^2 n!}{i!(n-i)!}=\sum_{i=1}^{n}\binom{n}{i}i^2=n(n+1)2^{n-2}$

Hope it will help you!