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I'm trying to find a way to represent $\sum_{n=1}^\infty n^2$ as a partial sum. I know that every term in this series can be represented, for example when $n=5$, as $5^2+4^2+3^2+2^2+1^2$. I know that $1+2+3+4...$ can be represented as $\frac{n(n+1)}{2}$ but I can't figure out how to add the squared terms to each number.

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    $\begingroup$ are you searching for $$\sum_{i=1}^ni^2$$? $\endgroup$ Nov 26 '17 at 6:30
  • $\begingroup$ Yeah I forgot to add the sigma notation $\endgroup$
    – John P
    Nov 26 '17 at 6:32
  • $\begingroup$ What does the $\infty$ do in your sum then? $\endgroup$
    – user499203
    Nov 26 '17 at 6:34
  • $\begingroup$ see here math.stackexchange.com/questions/183316/… $\endgroup$ Nov 26 '17 at 6:37
  • $\begingroup$ $f(N) =\sum_{n=1}^N n^2$ is a polynomial of degree $3$, and it is not hard to find its coefficients, the largest first : If $f(N) = \sum_{d=0}^3 c_d N^d$ then $f(N+1)-f(N) = \sum_{d=0}^3 c_d ((N+1)^d-N^d)$ $ = c_1 +c_2 (N^2+2N+1-N^2) + c_3 (N^3+3N+3N^2+1-N^3))$. Can you solve for $c_1,c_2,c_3$ to obtain $f(N+1)-f(N) = N^2$ ? $\endgroup$
    – reuns
    Nov 26 '17 at 6:43
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What you're looking for is $$\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}$$

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It is $$\frac{n(n+1)(2n+1)}{6}$$

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Using finite calculus we can write $n^2=n^\underline 2+n$ where $n^\underline k$ represent a falling factorial. Thus

$$\sum_{n=k}^m n^2=\sum_{n=k}^m (n^\underline 2+ n)=\sum\nolimits_k^{m+1}(n^\underline 2+n)\,\delta n=\left[\frac{n^\underline 3}3+\frac{n^\underline 2}2\right]_{n=k}^{n=m+1}\\=\frac16(2(m+1)^\underline 3+3(m+1)^\underline 2-2k^\underline 3-3k^\underline 2)$$

For $k=1$ it reduces to $\frac16(m+1)^\underline 2(2m+1)$.

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It is $\sum_{i=0}^n i^2=\frac{n(n+1)(2n+1)}{6}$.

There are some ways do figure that out. You might calculate the first cases $n=1,2,3,4,5$ and try to to find the formula. This is arguably a little bit tough.

That this formula holds can be proven by induction and is easy.

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    $\begingroup$ "There are some ways do figure that out." - There is actually a way to turn this problem into a system of linear equations. For any number $k\in\mathbb N$, finding $\sum_{i=1}^ni^k$ is equivalent to solving a linear system of $k+3$ equations. $\endgroup$
    – user499203
    Nov 26 '17 at 6:55

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