# how do I derive $1 + 4 + 9 + \cdots + n^2 = \frac{n (n + 1) (2n + 1)} 6$ [duplicate]

Possible Duplicate:
Proof that $\sum\limits_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$?

I am introducing my daughter to calculus/integration by approximating the area under y = f(x*x) by calculating small rectangles below the curve.

This is very intuitive and I think she understands the concept however what I need now is an intuitive way to arrive at $\frac{n (n + 1) (2n + 1)} 6$ when I start from $1 + 4 + 9 + \cdots + n^2$.

In other words, just how came the first ancient mathematician up with this formula - what were the first steps leading to this equation? That is what I am interested in, not the actual proof (that would be the second step).

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## marked as duplicate by Chris Taylor, Hans Lundmark, Américo Tavares, Fabian, Stefan HansenJan 24 '13 at 18:52

Induction is elementary too. How old is she? –  hjpotter92 Jan 24 '13 at 17:34
she's 16 but math is not exactly easy for her. So I am looking for something intuitive. –  nn4l Jan 24 '13 at 17:36
This question was asked a multiple times, so have a look here: math.stackexchange.com/questions/60858/…, math.stackexchange.com/questions/282349/… –  Alex Jan 24 '13 at 17:37
I don't think this counts as a duplicate, since its asking for a proof which isn't by induction. –  dinoboy Jan 24 '13 at 17:41
This is almost a duplicate of Proof that $\sum\limits_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$?. –  Américo Tavares Jan 24 '13 at 17:45

Same as you can prove sum of n = n(n+1)/2 by

*oooo
**ooo
***oo
****o


you can prove $\frac{n (n + 1) (2n + 1)} 6$ by building a box out of 6 pyramids:

Sorry the diagram is not great (someone can edit if they know how to make a nicer one). If you just build 6 pyramids you can easily make the n x n+1 x 2n+1 box out of it.

• make 6 pyramids (1 pyramid = $1 + 2^2 + 3^2 + 4^2 + ...$ blocks)
• try to build a box out of them
• measure the lengths and count how many you used.. that gives you the formula

Using these (glued)

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this could be a fun activity to let children discover the formula by themselves. –  user58512 Jan 24 '13 at 18:12
someone knew this already math.stackexchange.com/a/48152/58512 –  user58512 Jan 24 '13 at 18:35
this is very much what I was looking for. An intuitive way of understanding the series. –  nn4l Jan 24 '13 at 20:40

Try using $(n+1)^3-n^3=3n^2+3n+1$. Take the sum of both sides.

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More detail here. –  Chris Taylor Jan 24 '13 at 18:20

HINT: Use the fact that $(k+1)^3=k^3+3k^2+3k+1$

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This isn't really an answer. –  Chris Taylor Jan 24 '13 at 18:27
@ChrisTaylor: sure, it's not. It's a HINT. –  Chris's sis Jan 24 '13 at 18:28
Yes - I was wondering why you would give a hint when it would be just as easy to give an answer! –  Chris Taylor Jan 24 '13 at 18:34
@ChrisTaylor: giving a complete answer is often an easy thing, but giving help to OP and let him think such that he/she could understand the problem might be harder, but better for him/her. –  Chris's sis Jan 24 '13 at 18:45
Basically you put 3 $n \times n$ squares onto a big rectangle. On the left you get a side of $2n+1$ while the top has length $1 + 2 + ... + n = n(n+1)/2$. Thus $3 \cdot (1^2 + ... + n^2) = n(n+1)(2n+1)/2$, which gives the desired result. –  dinoboy Jan 24 '13 at 21:24
hint:let $a_n=1+2^2+...+(n)^2$ then $a_n-a_{n-1}=n^2-(n-1)^2$ then use recursive relation to prove $a_n= \frac{n (n + 1) (2n + 1)} {6}$ totally let $a_n=\sum_{j=1}^\infty c_ja_{n-j} +f(n)$ in your question $f(n)=An^k$ then compute $a_n^p$ and$a_n^g$ then $a_n=a_n^p+a_n^g$