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Prove algebraically that $n(n+1)2^{n-2}=\sum_{k=1}^{n}k^2\dbinom{n}{k}$ where $n$ and $k$ are integers.

I start with the fact that $(x+1)^n=\sum_{k=0}^{n}\dbinom{n}{k}x^k$ and differentiate both sides of the equation twice, with respect to $x$.

The second derivative (with respect to $x$) of the above equation yields: $n(n-1)(x+1)^{n-2}=\sum_{k=0}^{n}k^2-k\dbinom{n}{k}x^{k-2}$.

Now, I am not sure how to proceed. I know I'm supposed to set $x = 1$. But even then, I am missing the $(n+1)$ on the left hand side and there is still a $-k$ on the right hand side.

Any help would be appreciated. Thank you.

Jones
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    Notice that to obtain $k^2$ you can differentiate once, multiply an $x$ and then differentiate again. – Phicar Feb 01 '21 at 23:39

2 Answers2

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Using your idea of differentiating the binomial formula & then setting $x=1$ ... gives\begin{eqnarray*} n2^{n-1} &=& \sum_{k=0}^{n} k \binom{n}{k} \\ n(n-1)2^{n-2}&=& \sum_{k=0}^{n} k(k-1) \binom{n}{k} \\ \end{eqnarray*} now add these & you will have your result.

Donald Splutterwit
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Hint: $$\begin{align} \sum_{k=1}^{n}k^2\dbinom{n}{k}&=\sum_{k=1}^{n}k^2\frac nk \dbinom{n-1}{k-1}\\ &=n\sum_{k=1}^{n}k\left[\dbinom{n}{k}-\dbinom{n-1}{k}\right]\\ &=n\sum_{k=1}^{n}\left[n\dbinom{n-1}{k-1}-(n-1)\dbinom{n-2}{k-1}\right]. \end{align}$$

user
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