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Possible Duplicate:
Proof that $\sum\limits_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$?
Summation of natural number set with power of $m$
How to get to the formula for the sum of squares of first n numbers?

how can one find the value of the expression, $(1^2+2^2+3^2+\cdots+n^2)$

Let,

$T_{2}(n)=1^2+2^2+3^2+\cdots+n^2$

$T_{2}(n)=(1^2+n^2)+(2^2+(n-1)^2)+\cdots$

$T_{2}(n)=((n+1)^2-2(1)(n))+((n+1)^2-2(2)(n-1))+\cdots$

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HOLYBIBLETHE
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  • @RossMillikan but that duplicate doesn't really provide much help for the OP. –  Aug 30 '12 at 00:14
  • I'm trying to locate another question that deals specifically with $1^2 + 2^2 \dots + n^2$. Search sucks! –  Aug 30 '12 at 00:16
  • @JenniferDylan: Robert Israel's answer provides a nice link for powers besides 2 – Ross Millikan Aug 30 '12 at 00:16
  • @RossMillikan may be you're right. I was quick in my judgement. I just want us to slow down on the closures a bit. –  Aug 30 '12 at 00:18
  • @JenniferDylan: I am pretty sure this has been asked before, I just can't find it. Also, thanks for providing the link on that other question, I added them to the list of generalizations of common questions because they are asked so often. – Eric Naslund Aug 30 '12 at 00:32
  • Related: http://math.stackexchange.com/questions/95047/combinatorial-interpretation-of-sum-of-squares-cubes and http://math.stackexchange.com/questions/122546/gaussian-proof-for-the-sum-of-squares and http://math.stackexchange.com/questions/65861/summation-of-natural-number-set-with-power-of-m – Eric Naslund Aug 30 '12 at 00:36
  • Also How do I come up with a function to count a pyramid of apples?
    Proof that $\sum\limits_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$?
    Finite Sum of Power?     – MJD Aug 30 '12 at 00:55
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    I downvoted because I thought Rajesh should have caught a clue after his sum-of-cubes question was closed an hour before. – MJD Aug 30 '12 at 01:05
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    @Jennifer I am sympathetic to the argument that we too often close questions as "duplicates" when they are not, and that closers should take more care to make sure that the similar questions are really close. But in this case I think there are not only one but several exact duplicates. – MJD Aug 30 '12 at 01:33
  • @JenniferDylan, your memory is good, it was asked before on http://math.stackexchange.com/questions/48080/proof-that-sum-limits-k-1nk2-fracnn12n16 – NoChance Aug 30 '12 at 02:55

4 Answers4

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In general, $$\sum_{i=1}^{n}i^{2} = \frac{n(n+1)(2n+1)}{6}.$$

A collection of proofs of this fact can be found here.

Kirk Boyer
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Hint: $(n+1)^3-n^3=3n^2+3n+1$ and use telescopic sum.

Aang
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The claim is that

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

We will verify this by induction.

Clearly $n = 1$ holds.

Suppose the formula holds for $n$. Lets verify it holds for $n + 1$.

$$\sum_{i = 1}^{n + 1} i^2 = \sum_{i = 1}^n i^2 + (n + 1)^2 = \frac{n(n + 1)(2n + 1)}{6} + (n + 1)^2 \\ = \frac{n(n + 1)(2n + 1)}{6} + \frac{6(n^2 + 2n + 1)}{6} \\ = \frac{2n^3 + 3n^2 + n}{6} + \frac{6n^2 + 12 n + 6}{6} \\ = \frac{2n^3 + 9n^2 + 13n + 6}{6}$$

If you factor you get

$$= \frac{(n + 1)(n + 2)(2n + 3)}{6} = \frac{(n + 1)((n + 1) + 1)(2(n + 1) + 1)}{6}$$

The result follows for $n + 1$. So by induction the formula holds.

William
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If you can find or sketch some 3D blocks, there is a fun geometric proof.

Fix some $n$. If you are doing this with real blocks, $n=3$ or $4$ should be convincing. Let's take $n=4$ for now.

Make a $4\times 4\times 1$ base, laid flat, which of course has volume $4^2$.

Now make a $3\times 3 \times 1$ brick and place it, laid flat, above the base with some corners aligned. Continue in this way up to the top, making smaller squares and always aligning with the same corner. You now have a 3D corner of stairs whose volume is $1^2+\cdots +n^2$.

Now the fun part. Make 5 more of these "stair corners", for a total of 6. These six toys can be turned sideways and upside down, and then pieced together to make an $n\times(n+1)\times(2n+1)$ rectangular solid.

2'5 9'2
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