Compute the following sum or give me a guidance (I try to solve it by use of logarithm but I could not ) $$\sum_{i=1}^{n}i^22^i=?$$
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Some related older posts: http://math.stackexchange.com/questions/338852/find-a-closed-form-of-the-series-sum-n-0-infty-n2xn http://math.stackexchange.com/questions/36008/summation-by-parts-of-sum-k-0nk22k http://math.stackexchange.com/questions/460470/calculate-sum-limits-i-0n-i2-cdot-2i http://math.stackexchange.com/questions/912801/induction-proof-for-sum-limits-i-1n-i22i-n22n1-n2n23-cdot2 – Martin Sleziak Oct 09 '16 at 09:32
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Hint: try differentiation under summation,
$$ \sum_i i^2 2^i = \sum_i i^2 2^{i x} |_{x=1}, $$
with
$$ i^2 2^{i x} = \frac{1}{(\log 2)^2} \frac{\partial^2}{\partial x^2} 2^{i x}. $$
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You are looking for $$S_n=\sum_{i=1}^{n}i^2x^i$$ Rewrite it as $$S_n=\sum_{i=1}^{n}(i(i-1)+i)x^i=\sum_{i=1}^{n}i(i-1)x^i+\sum_{i=1}^{n}ix^i$$ $$S_n=x^2\sum_{i=1}^{n}i(i-1)x^{i-2}+x\sum_{i=1}^{n}ix^{i-1}$$ $$S_n=x^2 \left( \sum_{i=1}^{n}x^{i}\right)''+x\left( \sum_{i=1}^{n}x^{i}\right)'$$ I am sure that you can take it from here. When finished, set $x=2$.
Claude Leibovici
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