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The sum of the squares of the first n natural numbers

Posted November 27th, 2007 by Isoscel

What is the sum of the squares of the first n natural numbers?

The answer is:

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

Here is a calculator that calculates this function for you:

n:

$ 1^2+2^2+...+n^2 $:

Proof:

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

We write the above identity for k from 1 to n:

$ 2^3=1^3+3\cdot 1^2+3\cdot 1+1 $
$ 3^3=2^3+3\cdot 2^2+3\cdot 2+1 $
$ 4^3=3^3+3\cdot 3^2+3\cdot 3+1 $
$ 5^3=4^3+3\cdot 4^2+3\cdot 4+1 $
...
$ (n+1)^3=n^3+3\cdot n^2+3\cdot n+1 $

We add this n identities and we get:

$$(n+1)^3=1^3+3\sum_{k=1}^{n} k^2+3\sum_{k=1}^{n} k+\sum_{k=1}^{n}1$$
$$n^3+3n^2+3n+1=1+3\sum_{k=1}^{n} k^2+3\frac{n(n+1)}{2}+n$$
$$n^3+3n^2+3n=3\sum_{k=1}^{n} k^2+3\frac{n(n+1)}{2}+n$$
$$n^3+3n^2+2n - 3\frac{n(n+1)}{2}=3\sum_{k=1}^{n} k^2$$
$$n(n^2+3n+2) - 3\frac{n(n+1)}{2}=3\sum_{k=1}^{n} k^2$$
$$n(n+1)(n+2) - 3\frac{n(n+1)}{2}=3\sum_{k=1}^{n} k^2$$
$$n(n+1)(n+2-\frac{3}{2})=3\sum_{k=1}^{n} k^2$$
$$n(n+1)(n+2-\frac{3}{2})=3\sum_{k=1}^{n} k^2$$
$$n(n+1)(\frac{2n+1}{2})=3\sum_{k=1}^{n} k^2$$
$$\frac{n(n+1)(2n+1)}{6}=\sum_{k=1}^{n} k^2$$
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Comments

Proof of the Sum of squares posted nov. 27, 2007 by Isocel

On May 22nd, 2011 Andreas says:

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I followed very carefully your proof and think it's all right.
Please explain the reasoning in this "adding the n identities" step:

(n+1)^3 = 1^3 + summation of k-squares+ summation of k + n

where does this 1 cube appear???

Greetings

Andreas

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about sum

On May 24th, 2011 Structure says:

You have
We write the above identity for k from 1 to n:

$ 2^3=1^3+3\cdot 1^2+3\cdot 1+1 $
$ 3^3=2^3+3\cdot 2^2+3\cdot 2+1 $
$ 4^3=3^3+3\cdot 3^2+3\cdot 3+1 $
$ 5^3=4^3+3\cdot 4^2+3\cdot 4+1 $
...
$ (n+1)^3=n^3+3\cdot n^2+3\cdot n+1 $
so

$$(2^3+3^3+4^3+...+n^3)+(n+1)^3=1^3+(2^3+3^3+4^3+...+n^3)+3\sum k^2+3\sum k+(1+1+....+1)$$

Now you can reduce $ (2^3+3^3+4^3+...+n^3) $ but the first $ 1^3 $ remains.

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