Some parts of the text might not be clear so please ask about them in the comments. Sorry for the uploaded image as it was taking a long time for me to write the whole proof.
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11+1 for the photo – Jochen Oct 24 '20 at 15:16
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2Please type your attempts here. Do not attach images. – Célio Augusto Oct 24 '20 at 15:17
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2Lol for the pic – user577215664 Oct 24 '20 at 15:21
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1@Jochen Me too :) – Saša Oct 24 '20 at 15:23
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3It's an original version of a Brillant Mind when Nash does maths on the window... – user577215664 Oct 24 '20 at 15:25
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2@CélioAugusto His attempt is better documented than my birth date in state papers. Leave him alone. And the question is interesting too. However, whenever I try to check his attempt, I stop at his legs and cannot move. :) – Saša Oct 24 '20 at 15:26
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1very good job!!! Its a bit long but it seems truly original – Masacroso Oct 24 '20 at 15:29
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1This webpage might help. – Oct 24 '20 at 16:05
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1Different Methods to compute the sum $\displaystyle\sum_{k = 1}{1 \over k^{2}}$ has, for the time being, $\displaystyle\large 55$ answers !!!. – Felix Marin Oct 24 '20 at 16:07
1 Answers
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The usual short proof is$$\begin{align}\sum_{n\ge0}x^n&=\frac{1}{1-x}\\\implies\sum_{n\ge0}n^2x^n&=x\frac{d}{dx}\left(x\frac{d}{dx}\sum_{n\ge0}x^n\right)\\&=x\frac{d}{dx}\left(\frac{1}{(1-x)^2}-\frac{1}{1-x}\right)\\&=\frac{x(1+x)}{(1-x)^3}\end{align}$$with $x=\tfrac12$. It looks like your approach is more similar to a different one that can also be made reasonably concise:$$\sum_{n\ge0}n^2x^n=\sum_{n\ge0}\left(2\binom{n}{2}+n\right)x^n=\left(x^2\frac{d^2}{dx^2}+x\frac{d}{dx}\right)\sum_{n\ge0}x^n=\frac{2x^2}{(1-x)^3}+\frac{x}{(1-x)^2}.$$
J.G.
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