Binomial Theorem and $n(n+1)2^{n-2} = \sum_{i=1}^ni^2\binom{n}{i}$

Question

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

I had proved this combinatorially but also trying to derive this identity using binomial theorem.

From the bionomial theorem, one could easily get $2^n = \sum_{k=0}^n\binom{n}{k}$

Using this LHS could be equated with$\binom{n+1}{2}{1\over2} \sum_{k=0}^n\binom{n}{k}$

I'd like to go further to make this form closer to $\sum_{i=1}^ni^2$ but it's hard to imagine where to go.

Any advice?

Answer 1

You already have $$(x+1)^n = \sum_{i=1}^n x^i{n\choose i},$$ $$n(x+1)^{n-1} = \sum_{i=1}^n i{n\choose i} x^{i-1}$$ $$nx(x+1)^{n-1} = \sum_{i=1}^n i{n\choose i} x^i$$ $$n(x+1)^{n-1} + n(n-1)x(x+1)^{n-2} = \sum_{i=1}^n i^2{n\choose i} x^{i-1}.$$

Now, taking $x=1$, you have $$n2^{n-1}+ n(n-1)2^{n-2} = \sum_{i=1}^ni^2{n \choose i}.$$

Answer 2

Let, $$(1+x)^n=\binom{n}{0}+\binom{n}{1}x+\binom{n}{2} x^2\cdots +\binom{n}{n} x^n$$$$$$ Differentiating wrt x, $$n(1+x)^{n-1}=\binom{n}{1}+2\binom{n}{2}x+3\binom{n}{3} x^2+\cdots +n\binom{n}{n} x^{n-1}$$$$$$ Multiplying both sides by x, $$n(1+x)^{n-1} x=\binom{n}{1}x+2\binom{n}{2}x^2+3\binom{n}{3}x^3+\cdots +n\binom{n}{n}x^n$$$$$$ Differential wrt x, $$\frac{d}{dx}(n(1+x)^{n-1}x)=\binom{n}{1}+2^2\binom{n}{2}x+3^2\binom{n}{3}x^2+\cdots +n^2\binom{n}{n} x^{n-1}$$$$$$ $$n((n-1)x(1+x)^{n-2}+(1+x)^{n-1})=\binom{n}{1}+2^2\binom{n}{2}x+3^2\binom{n}{3}x^2+\cdots +n^2\binom{n}{n} x^{n-1}$$$$$$ $$n(1+x)^{n-2}(nx+1)=\binom{n}{1}+2^2\binom{n}{2}x+3^2\binom{n}{3}x^2+\cdots +n^2\binom{n}{n} x^{n-1}$$$$$$ Put $x=1$, $$n(1+1)^{n-2}(n+1)=\binom{n}{1}+2^2\binom{n}{2}+3^2\binom{n}{3}+\cdots +n^2\binom{n}{n}$$$$$$ $$n(n+1)2^{n-2}=\binom{n}{1}+2^2\binom{n}{2}+3^2\binom{n}{3}+\cdots +n^2\binom{n}{n}$$ Hence... $$n(n+1)2^{n-2} = \sum_{i=1}^ni^2\binom{n}{i}$$

Answer 3

For $i\ge0,$ $$i^2\binom ni=i(i-1)\binom ni+i\binom ni$$

Now for $i\ge1,$

$$i\binom ni=i\cdot\dfrac{n\cdot(n-1)!}{i\cdot(i-1)!\{n-1-(i-1)\}!}=n\binom{n-1}{i-1}$$

and for $i\ge2,$ $$i(i-1)\binom ni=i(i-1)\cdot\dfrac{n(n-1)\cdot(n-2)!}{i(i-1)\cdot(i-2)!\{n-2-(i-2)\}!}=n(n-1)\binom{n-2}{i-2}$$

$$\implies \sum_{i=1}^ni^2\binom ni=\binom n1+n\sum_{i=2}^n\binom{n-1}{i-1}+n(n-1)\sum_{i=2}^n\binom{n-2}{i-2}$$

Now $\displaystyle\sum_{i=2}^n\binom{n-1}{i-1}=-\binom{n-1}0+\sum_{i=1}^n\binom{n-1}{i-1}$

$\displaystyle=-1+\sum_{j=0}^{n-1}\binom{n-1}j\ \ \ \ (1)$ (setting $i-1=j$)

and $\displaystyle\sum_{i=2}^n\binom{n-2}{i-2}=\sum_{k=0}^{n-2}\binom{n-2}k\ \ \ \ (2)$ (setting $i-2=k$)

Now for integer $m\ge0,$ $$(a+b)^m=\sum_{r=0}^m\binom mra^{m-r}b^r$$

Set $a=b=1$ to find $$(1+1)^m=\sum_{r=0}^m\binom mr$$

Set $m=n-1, n-2$ for $(1),(2)$ respectively.