I'm asked to prove the identity $$n(n+1)2^{n-2} = \sum_{k=1}^{n}k^2\binom{n}{k}$$ using the binomial theorem.

What I've been able to come up with so far is letting $$f(x) = (x+1)^n = \sum_{k=0}^{n}x^k\binom{n}{k}$$ taking two derivatives gives $$f''(x) = n(n-1)(x+1)^{n-2} = \sum_{k=2}^{n}k(k-1)x^{k-2}\binom{n}{k}$$ If I plug in x = 1, I get something super close to the required identity: $$n(n-1)2^{n-2} = \sum_{k=2}^{n}k(k-1)\binom{n}{k}$$

I think the method I'm using is correct, and I just need to tweak it a little bit... Any help is appreciated!

  • $\begingroup$ Easy. Just multiply through by $x$ before you differentiate the second time! $\endgroup$ – almagest Mar 21 '16 at 17:44

After you take the first derivative, you have


Now multiply both sides by $x$ before you differentiate again.

Here’s an alternative that uses the binomial theorem only to evaluate $\sum_k\binom{m}k=2^m$. Using the fact that $k\binom{n}k=n\binom{n-1}{k-1}$, we can calculate

$$\begin{align*} \sum_kk^2\binom{n}k&=\sum_kkn\binom{n-1}{k-1}\\ &=n\sum_kk\binom{n-1}{k-1}\\ &=n\sum_k(k+1)\binom{n-1}k\\ &=n\sum_kk\binom{n-1}k+n\sum_k\binom{n-1}k\\ &=n\sum_k(n-1)\binom{n-2}{k-1}+n2^{n-1}\;, \end{align*}$$

and I expect that you can finish it from there.

  • $\begingroup$ Multiplying both sides of the first derivative by $x$ gives me $$xn(x+1)^{n-1} = \sum_{k=1}^{n}kx^{k}\binom{n}{k}$$ Differentiating gives $$f''(x) = xn(n-1)(x+1)^{n-2} = \sum_{k=1}^{n}k^2x^{k-1}\binom{n}{k}$$ Plugging in $x=1$: $$n(n-1)2^{n-2} = \sum_{k=1}^{n}k^2\binom{n}{k}$$ Which still contains an $n-1$ term where I need a $n+1$ term, what am I doing wrong here? $\endgroup$ – Jim Jj Mar 21 '16 at 19:15
  • $\begingroup$ @Jim: Your differentiation is wrong: $x(x+1)^{n-1}$ is a product, so you’ll have two terms in the derivative (before simplification). $\endgroup$ – Brian M. Scott Mar 21 '16 at 19:17

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