Prove that $\sum_{k=0}^n \binom{n}{k} k^2 = n(n+1) 2^{n-2}$

Question

$$\sum_{k=0}^n \binom{n}{k} k^2 = n(n+1) 2^{n-2}$$

Answer 1

HINT: You know $$f(x):=\sum_{k=0}^n \binom{n}{k}x^k = (1+x)^n.$$ Compute $f'(x)$ and $f''(x)$ using both the sum and the power expressions. Asuitable linear combination fo $f(1)$, $f'(1)$, $f''(1)$ produces the sum in question.

Answer 2

If you want to use induction, prove in the following order (using Pascal's formula) that:

• $\displaystyle\sum_{k=0}^n \binom{n}{k}= 2^{n}$,
• $\displaystyle\sum_{k=0}^n k\binom{n}{k}= n2^{n-1}$,
• $\displaystyle\sum_{k=0}^n k^2\binom{n}{k}= n(n+1)2^{n-2}$.

For a complete proof see this.

Answer 3

For $k\ge 2,$ $$k^2\binom nk=\{k(k-1)+k\}\binom nk$$ $$=k(k-1)n(n-1)\frac{(n-2)!}{(k-2)!k(k-1)\{(n-2)-(k-2)\}!}+kn\frac{(n-1)!}{(k-1)! k \{(n-1)-(k-1)\}}$$ $$=n(n-1)\binom{n-2}{k-2}+n\binom{n-1}{k-1}$$

$$\sum_{0\le k\le n}k^2\binom nk$$ $$=n(n-1)\sum_{2\le k\le n}\binom{n-2}{k-2}+n\sum_{1\le k\le n}\binom{n-1}{k-1}$$ $$=n(n-1)(1+1)^{n-2}+n(1+1)^{n-1}$$