1. ## sum of numbers

Hi My Friends , What's up ?

The sum of squares :

1^2 + 2^2 + 3^2 + ... + n^2= n(n+1)(2n+1)/6

I have two questions :

1- how do we derived the equation above ??

2- what is the equation for the following sum :

(1^0.25)+(2^0.25)+(3^0.25)+ ... +(n^0.25)

Thank you very much

2. Originally Posted by metallica007
Hi My Friends , What's up ?

The sum of squares :

1^2 + 2^2 + 3^2 + ... + n^2= n(n+1)(2n+1)/6

I have two questions :

1- how do we derived the equation above ??
There are lots of ways of deriving the formula for the sum $S_n$ of the squares of the first $n$ positive integers (see the PlanetMath article). One way is to observe that the sum must be a cubic in $n$ (this is obviously the case as the third differences of the sequence $\{S_n, n=1, 2, ..\}$ are constant). Then we can use the first few terms of the sequence to find the coefficients of the cubic.

2- what is the equation for the following sum :

(1^0.25)+(2^0.25)+(3^0.25)+ ... +(n^0.25)
I believe there is no elementary closed form for this (there is a formula in terms of generalised harmonic numbers or the polylogarithm function but these are not elementary).

CB

3. There are various ways to solve that problem. An interesting one is to use geometric series:

Let $S_n = 1^2 + 2^2 + \cdot\cdot\cdot + n^2$

Let $S = 1 + r + r^2 + \cdot\cdot\cdot + r^{n}$
$\implies \frac{dS}{dr} = 1 + 2r +3r^2 + \cdot\cdot\cdot + nr^{n-1}$
$\implies r\frac{dS}{dr} = r + 2r^2 +3r^3 + \cdot\cdot\cdot + nr^{n}$
$\implies \frac{d}{dr}\left(r\frac{dS}{dr}\right) = 1 + 2^2r +3^2r^2 + \cdot\cdot\cdot + n^2r^{n-1}$

Therefore $S_n = \frac{d}{dr}\left(r\frac{dS}{dr}\right)\bigg{|}_{r =1}$

To evaluate the right hand side, use the fact that $S = \frac{r^{n+1} - 1}{r - 1}$

4. Thank you very much my friends

5. Code:
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