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$$\sum_{k=1}^{n}{k^2}\cdot\dbinom{n}{k} = n(n+1)2^{n-2}$$

Attempt: I have re-written in it as follows: $$\sum_{k=1}^{n}\dbinom{k}{1}\dbinom{k}{1}\dbinom{n}{k} = \dbinom{n}{1}\dbinom{n+1}{1} 2^{n-2}$$

the LHS seems like a case where a committee of k people is to be formed from n people. We choose a president in k ways. Then, we choose a secretary in k ways again. Which means that the president could be chosen as a secretary as well? (probably not the best example). How can I use the RHS to show the same thing?

RobPratt
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D_M
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1 Answers1

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Algebriac Proof:

Rewrite $k^2=k(k-1)+k$. We have \begin{eqnarray*} \sum_{k=1}^{n} k^2 \binom{n}{k} &=& \sum_{k=1}^{n} k(k-1) \binom{n}{k} + \sum_{k=1}^{n} k \binom{n}{k} \\ &=& n(n-1) \sum_{k=2}^{n} \binom{n-2}{k-2} + n \sum_{k=1}^{n} \binom{n-1}{k-1} \\ &=& n(n-1) 2^{n-2} + n 2^{n-1} = n(n+1)2^{n-2}.\\ \end{eqnarray*}

Combinatorial Proof:

We choose a committee of size $k$ from $n$ people and then choose a president & secretary (possible the same person). This is the LHS

For the RHS there are $2$ case.

$1)$ Choose the president ($n$ ways) and secretary ($n-1$ ways) to be a different from the president & then the rest of the committee ($2^{n-2}$ ways.)

$2)$ Choose the president & secretary to be the same person ($n$ ways) and the rest of the committee ($2^{n-1}$ ways.)

Now add these to get the result.

Donald Splutterwit
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