How to prove $\sum_{r=1}^{n} r^{2}\binom {n} {r} = n(n+1)2^{n-2}$?

Question

How to prove $$\sum_{r=1}^{n} r^{2}\binom {n} {r} = n(n+1)2^{n-2}$$ By using binomial theorem?

For a combinatorial proof, see https://math.stackexchange.com/questions/1923215/give-a-combinatorial-proof-nn12n-2-sum-k-1nk2-binomnk/1923224#1923224 . Through the binomial theorem, it is enough to apply $(xD)^2$ to both sides of $$\sum_{r=0}^{n}\binom{n}{r}x^r = (1+x)^n,$$ then evaluate at $x=1$. — Jack D'Aurizio, Jan 18 '18 at 17:44
This is not a duplicate. The OP is explicilty asking for use of the binomial theorem, while the duplicate emphasizes using combinatorial arguments. — Martin Argerami, Jan 18 '18 at 18:55

score 3 · Answer 1 · answered Jan 18 '18 at 17:46

You have $$ \sum_{r=0}^n{n\choose r}x^r=(1+x)^n. $$ If you differentiate, $$\tag1 \sum_{r=1}^nr{n\choose r}x^{r-1}=n(1+x)^{n-1}.$$ Differentiate again: $$\tag2 \sum_{r=1}^nr(r-1){n\choose r}x^{r-2}=n(n-1)(1+x)^{n-2}. $$ If you evaluate $(1)$ at $x=1$, $$\tag3 \sum_{r=1}^nr{n\choose r}=n\,2^{n-1} $$ If you evaluate $(2)$ at $x=1$, $$\tag4 \sum_{r=1}^nr(r-1){n\choose r}=n(n-1)\,2^{n-2}. $$ If you add $(3)$ and $(4)$, $$ \tag5 \sum_{r=1}^nr^2{n\choose r}=n\,2^{n-1}+n(n-1)2^{n-2}=(2n+n(n-1))\,2^{n-2}=n(n+1)\,2^{n-2}. $$

score 1 · Answer 2 · answered Jan 18 '18 at 17:51

$$\sum_{r=1}^n r^2\binom {n}{r}=n\sum_{r=1}^n r \binom {n-1}{r-1}$$ $$=n\sum_{r=1}^n (r-1+1)\binom {n-1}{r-1}$$ $$=n\left[\sum_{r=1}^n (r-1)\binom {n-1}{r-1} + \sum_{r=1}^n \binom {n-1}{r-1}\right]$$ $$=n\left[(n-1) \sum_{r=2}^n \binom {n-2}{r-2}+ \sum_{r=1}^n \binom {n-1}{r-1}\right]$$ $$=n\left[(n-1).2^{n-2} + 2^{n-1}\right]$$ $$=n(n+1).2^{n-2}$$

How to prove $\sum_{r=1}^{n} r^{2}\binom {n} {r} = n(n+1)2^{n-2}$?

2 Answers2