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This question already has an answer here:

Is there any way to prove that: $$ 1^2+2^2+3^2+\ldots +n^2=\frac{n(n+1)(2n+1)}{6}$$ but WITHOUT using mathematical induction? (I don't know, maybe through some creative graphical demonstration?)

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marked as duplicate by PSPACEhard, Dietrich Burde, Juniven, projectilemotion, Simply Beautiful Art Mar 24 '17 at 11:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ $f(n) = \frac{n(n+1)(2n+1)}{6}, f(n)-f(n-1) = n^2, f(0) = 0$ $\endgroup$ – reuns Mar 24 '17 at 11:08
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Yes, there is a graphic proof by Man-Keung Siu, appeared in: Mathematics Magazine March, 1984

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  • $\begingroup$ This is very elegant. $\endgroup$ – mlc Mar 24 '17 at 11:07
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The answer is a polynomial of the third degree because the first order finite difference of a polynomial is a polynomial of one degree lower.

This polynomial is obtained by Lagrangian interpolation on the points $(0,0),(1,1),(2,5)$ and $(3,14)$. As there is no independent term, you make is slightly easier by interpolating $P(n)/n$ on $(1,1),(2,5/2)$ and $(3,14/3)$.

[https://www.wolframalpha.com/input/?i=interpolate+(0,0)+(1,1)+(2,5)+(3,14), https://www.wolframalpha.com/input/?i=interpolate++(1,1)+(2,5%2F2)+(3,14%2F3) ]

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