Summation identity

**zhangvict** · July 17th 2011, 02:04 PM

Why is the following identity true?

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

**Siron** · July 17th 2011, 02:07 PM

There's nothing to see? ...

**zhangvict** · July 17th 2011, 02:10 PM

sorry, had problems posting the equation. It is there now

**pickslides** · July 17th 2011, 02:26 PM

The best way to answer this 'why?' is to prove it yourself.

Do you know of mathematical induction?

**Also sprach Zarathustra** · July 17th 2011, 02:41 PM

Originally Posted by zhangvict

Why is the following identity true?

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

$\sum_{r=0}^{n}r^2=?$

Solution via 'Differences Series' (Or LaTex practice

)

Step no. 0:
$\sum_{r=0}^{n}r^2$

Step no. 1:
$\sum_{r=0}^{n+1}r^2-\sum_{r=0}^{n}r^2=n^2$

Step no. 2:
$n^2-(n-1)^2=2n-1$

Step no. 3:
$2n-1-[2(n-1)-1]=2$

Step no. 4:
$2-2=0$

Now we find the generating function from 4 and up...

The generating function of #4 is $x+x^2$

The generating function of #3 is $\frac{x+x^2}{1-x}$

The generating function of #2 is $\frac{x+x^2}{(1-x)^2}$

The generating function of #1 is $\frac{x+x^2}{(1-x)^3}$

The generating function of #0 is $\frac{x+x^2}{(1-x)^4}$

We get that generating function of $(\sum_{r=0}^{n}r^2)_{n=0}^{\infty} is \frac{x+x^2}{(1-x)^4}$ .

So, $\sum_{r=0}^{n}r^2$ is the coefficient of $x^n in \frac{x+x^2}{(1-x)^4}$ :

$\frac{x+x^2}{(1-x)^4}=(x+x^2)\cdot \frac{1}{(1-x)^4}=(x+x^2)\sum_{i=0}^{\infty}\binom{i+3}{3}x^i$

Hence the coefficient of $x^n$ is:

$\binom{n-1+3}{3}+\binom{n-2+3}{3}$

The conclusion is:

$\sum_{r=0}^{n}r^2=\binom{n+2}{3}+\binom{n+1}{3}$

**Archie Meade** · July 17th 2011, 03:11 PM

Originally Posted by zhangvict

Why is the following identity true?

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

There are a number of ways to show why the formula holds.

One easy method is to use the fact that there is a relationship
between the sum of positive natural numbers
and the sum of their squares....

$\sum_{r=1}^nr=\frac{n(n+1)}{2}$

as it is a very basic arithmetic series.

$1+2=3;\;\;\;1^2+2^2=3\left(\frac{5}{3}\right)$

$1+2+3=6;\;\;\;1^2+2^2+3^2=6\left(\frac{7}{3}\right )$

$1+2+3+4=10;\;\;\;\;1^2+2^2+3^2+4^2=10\left(\frac{9 }{3}\right)$

Hence, induction can prove without doubt that

$\sum_{r=1}^nr^2=\frac{2n+1}{3}\sum_{r=1}^nr$

**Prove It** · July 17th 2011, 09:14 PM

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