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I'm new and this is my first question (though I've been lurking). English is not my native language. Studying on my own.

I'm really interested in deriving the formula $1^{2} + 2^{2} + 3^{2} + \cdots+ n^{2} = \frac{n(n+1)(2n+1)}{6}$ using only the following:

Given: for $n$, $a$, $b$, $c$, $z$, $y$ positive integers

$1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}$ $ a^2 + b^2 + c^2 \cdots+ z^2 =(a + b + \cdots + z)^2 - 2ab - 2ac - \cdots - 2az - 2bc - \cdots -2bz - \cdots - 2$zy

I could write

$$1^{2} + 2^{2} + 3^{2} + \cdots+ n^{2} = [\frac{n(n+1)}{2}]^{2} - 2(1\cdot2) - 2(1\cdot3) - \cdots- 2(1\cdot n) - 2(2\cdot3) - \cdots - 2(2\cdot n) - \cdots -2(n-1)(n)$$

And then notice that:

$1^{2} + 2^{2} + 3^{2} + \cdots+ n^{2} = [\frac{n(n+1)}{2}]^{2} - 2[1(2 + 3 + \cdots+n)] - 2[2(3 + 4 + \cdots+ n)] - \cdots -2[(n-1)(n)]$

Which is

$1^{2} + 2^{2} + 3^{2} + \cdots n^{2} = [\frac{n(n+1)}{2}]^{2} - 2[\frac{n(n+1)}{2} -1] - 2[\frac{2n(n+1)}{2} - (1 +2)] -\cdots -2[\frac{(n-1)n(n+1)}{2} - (1 +2 + \cdots+n -1)]$

Simplifying to

$1^{2} + 2^{2} + 3^{2} + \cdots+ n^{2} = [\frac{n(n+1)}{2}]^{2} - 2[(1 + 2 +\cdots+ n-1)\frac{n(n+1)}{2} - 1(n-1) -2(n-2) - \cdots - (n-1)1]$

Edit (29/01): thought about it

$1^{2} + 2^{2} + 3^{2} + \cdots+ n^{2} = [\frac{n(n+1)}{2}]^{2} - 2[(1 + 2 +\cdots+ n-1)\frac{n(n+1)}{2} - 2(1 + 2 +\cdots n-1)]$

And then

$1^{2} + 2^{2} + 3^{2} + \cdots+ n^{2} = [\frac{n(n+1)}{2}]^{2} - 2[(\frac{(n-1)n}{2})(\frac{n(n+1)}{2}) - \frac{2n(n-1)}{2}]$

So

$1^{2} + 2^{2} + 3^{2} + \cdots+ n^{2} = [\frac{n(n+1)}{2}]^{2} - 2[\frac{n^{2}(n^{2}-1)}{2} - \frac{2n(n-1)}{2}]$

Then

$1^{2} + 2^{2} + 3^{2} + \cdots+ n^{2} = [\frac{n^{2}(n^{2}+2n+1)}{4}] - 4[\frac{n^{4}-3n^{2}-2n}{4}]$

But now

$1^{2} + 2^{2} + 3^{2} + \cdots+ n^{2} = \frac{n(-3n^{3}+2n^{2}-11n-2)}{4}$

Ok, now I know it's messed up, but still can't figure why

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  • 3
    $\begingroup$ The Wikipedia article on square pyramidal numbers gives one way the closed form may be derived. $\endgroup$ – augurar Jan 30 '15 at 0:49
  • $\begingroup$ Why not using finite difference to find $a_n=\sum_{k=1}^n k^2$ and then proving it by induction? $\endgroup$ – Vincenzo Oliva Feb 13 '15 at 18:57
  • $\begingroup$ One proof: $(n+1)^3=n^3+3n^2+3n+1=(n-1)^3+3(n^2+(n-1)^2)+3(n+(n-1))+2=\cdots=1+3(n^2+\cdots +1^2)+\frac{3n(n+1)}2+n$ then rearrange the ends. $\endgroup$ – Jaycob Coleman Apr 3 '15 at 23:30
  • $\begingroup$ By your way. We get: $\displaystyle S_n=\sum_{k=1}^n k^2=\left(\sum_{k=1}^n k\right)^2-2\sum_{1\le i<j\le n}ij=\left(\frac{n(n+1)}{2}\right)^2-2\sum_{j=2}^n\sum_{i=1}^{j-1}ij$ $\displaystyle \quad = \left(\frac{n(n+1)}{2}\right)^2-2\sum_{j=2}^nj\frac{(j-1)j}{2}$ $\displaystyle \quad =\left(\frac{n(n+1)}{2}\right)^2-\sum_{j=2}^n j^3+\sum_{j=2}^n j^2$ $\displaystyle \quad =\left(\frac{n(n+1)}{2}\right)^2-\sum_{j=1}^n j^3+S_n$ $\displaystyle \Rightarrow \sum_{k=1}^n k^3=\left(\frac{n(n+1)}{2}\right)^2$ :) $\endgroup$ – hxthanh May 11 '15 at 8:12
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The best way I have seen is to start with a difference of $(p+1)$-th powers, where $p$ is the power in the summation you want (I think this is what Jaycob Coleman is alluding to in his comment on 3 April). In our case:

$(k+1)^3 - k^3 = 3k^2 + 3k + 1$

Then apply this identity from $k=1$ to $k=n$, where $n$ is the upper limit of the summation. This is a telescoping series:

$\boxed{\begin{eqnarray*} 2^3 - 1^3 &= 3\cdot1^2 + 3\cdot1 + 1 \\ 3^3 - 2^3 &= 3\cdot2^2 + 3\cdot2 + 1 \\ &... \\ (n+1)^3 - n^3 &= 3\cdot{n^2} + 3\cdot{n} + 1 \\ \end{eqnarray*}}$

Summing up all terms (and cancelling on the left-hand side):

$(n+1)^3 - 1^3 = 3\displaystyle\sum\limits_{k=1}^{n}{k^2} + 3\displaystyle\sum\limits_{k=1}^{n}{k} + \displaystyle\sum\limits_{k=1}^{n}{1}$

Calling the sum we want $S$, and recognising the last two sums as $\dfrac{n^2+n}{2}$ and $n$ we have,

$3S + 3\dfrac{n^2+n}{2} + 3n = (n+1)^3 - 1 = n^3 + 3n^2 + 3n$

Solve this to get $\boxed{S = \frac{1}{3}n^3 + \frac{1}{2}n^2 + \frac{1}{6}n}$ which can be factored as $\boxed{S = \dfrac{n(n+1)(2n+1)}{6}}$

Remarks

The beauty of this approach is that you can derive the sum of the $p-th$ powers of $1,...,n$ with only knowledge of the sums of all lower powers of $1,...,n$ and some algebra.

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