I've been struggling with this sum for an while, pluging n+1 instead of n, knowing that \binom{n+1}{k} = \binom{n}{k-1} + \binom{n}{k} and after some manipulation i've found this sum.
2\sum_{k=1}^{n} k^2\binom{n}{k}
I coudn't see any way I could get out of here and I don't know how to start this proof without the property of the sum of binomial coefficients.
Thanks in advance.
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