How do I derive $1 + 4 + 9 + \cdots + n^2 = \frac{n (n + 1) (2n + 1)} 6$

Question

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).

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: http://math.stackexchange.com/questions/60858/proof-by-induction-sum-limits-i-1ni2-fracnn12n16?rq=1, http://math.stackexchange.com/questions/282349/summing-rr3-using-induction/282367#282367 — 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
@AméricoTavares I agree - and there are many elementary non-inductive proofs there. — Chris Taylor, Jan 24 '13 at 18:25
@nn4l You can also do it backwards, that is a very easy proof but extremely artificial: $\frac{(k+1)(k+2)(2k+3)}{6}-\frac{k(k+1)(2k+1)}{6}=k^2$. Then your sum becomes telescopic... — N. S., Jan 24 '13 at 20:44

score 11 · Answer 1 · answered Jan 24 '13 at 17:36

11

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

answered Jan 24 '13 at 17:36

Ishan Banerjee

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

score 8 · Accepted Answer · 2013-01-24T18:16:59.750

8

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:

enter image description here enter preformatted text here

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) enter image description here

edited Jan 24 '13 at 18:16

answered Jan 24 '13 at 18:09

this could be a fun activity to let children discover the formula by themselves. – Jan 24 '13 at 18:12
someone knew this already http://math.stackexchange.com/a/48152/58512 – Jan 24 '13 at 18:35
1

this is very much what I was looking for. An intuitive way of understanding the series. – nn4l Jan 24 '13 at 20:40

score 4 · Answer 3 · answered Jan 24 '13 at 17:36

4

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

answered Jan 24 '13 at 17:36

user 1591719

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This isn't really an answer. – Chris Taylor Jan 24 '13 at 18:27
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@ChrisTaylor: sure, it's not. It's a HINT. – user 1591719 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
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@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. – user 1591719 Jan 24 '13 at 18:45

score 4 · Answer 4 · answered Jan 24 '13 at 17:37

4

There is a nice proof without words here : http://www.usamts.org/About/U_Gallery.php Its a geometric proof of sorts which I guess could count as more intuitive than the somewhat contrived induction.

answered Jan 24 '13 at 17:37

dinoboy

2,894

I must admit that I do not understand the geometric proof of my problem at all. The other examples are brilliant though. – nn4l Jan 24 '13 at 20:42
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

M.H · Answer 5 · 2013-01-24T18:28:18.693

2

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 $a_n=a_n^p+a_n^g$

edited Jan 24 '13 at 18:28

answered Jan 24 '13 at 18:11

M.H

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How do I derive $1 + 4 + 9 + \cdots + n^2 = \frac{n (n + 1) (2n + 1)} 6$

5 Answers5