Show the method used to evaluate $\displaystyle\sum\limits_{k=1}^{12} {12\choose{k}}k^2$
The answer is $159744.$
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@Soke I wrote a computer program to find the answer. :P – Siddhartha Dec 08 '15 at 21:15
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@Thomas do you mean the sum to be indexed with $k$, and $k$ going up to $n$ (instead of 12)? – MT_ Dec 08 '15 at 21:15
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Ok, there was a typo which I fixed. I changed i to k. – Siddhartha Dec 08 '15 at 21:16
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\begin{align} f(x)&=(1+x)^n=\sum_{k=0}^n{n \choose k}x^k\\ f'(x)&=n(1+x)^{n-1}=\sum_{k=1}^n{n \choose k}kx^{k-1}\\ xf'(x)&=nx(1+x)^{n-1}=\sum_{k=1}^n{n \choose k}kx^{k}\\ (xf'(x))'&=n(1+x)^{n-1}+(n-1)n(1+x)^{n-2}=\sum_{k=1}^n{n \choose k}k^2x^{k-1}\\ x&\leftarrow 1\\ \therefore \sum_{k=1}^n{n \choose k}k^2&=n2^{n-1}+(n-1)n2^{n-2}=n(n+1)2^{n-2}\\ \end{align}
Kay K.
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For anyone else having a brain freeze understanding the last step (I stared at it way longer than I should) simply let $x=1$. – Ian Miller Dec 09 '15 at 01:33
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Thanks. I loved your answer but found myself scratching my head a bit. – Ian Miller Dec 09 '15 at 02:15
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Another approach:
$$\sum_{k=1}^{12} {12 \choose k} k^2 = \sum_{k=2}^{12} {12 \choose k} k(k-1) + \sum_{k=1}^{12} {12 \choose k} k \\ = \sum_{k=2}^{12} \frac{12!}{(k-2)!(12-k)!} + \sum_{k=1}^{12} \frac{12!}{(k-1)!(12-k)!} \\ = \sum_{k=2}^{12} \frac{12!}{(k-2)!(10-(k-2))!} + \sum_{k=1}^{12} \frac{12!}{(k-1)!(11-(k-1))!} \\ = 12 \cdot 11 \sum_{k=2}^{12} {10 \choose k-2} + 12 \sum_{k=1}^{12} {11 \choose k-1}.$$
Can you finish from here? Note that this approach generalizes straightforwardly (albeit tediously) to any sum of the form $\sum_{k=0}^N {n \choose k} k^\ell$.
Ian
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