# What is this formula I found?

1. Nov 4, 2004

### StatusX

I was trying to come up with a formula for the sums of powers of n from 1 to x (ie, x(x+1)/2 for the first power, x(x+1)(2x+1)/6 for the second, etc), and in the process, I found this pretty cool formula:

$$x^r = \sum_{n=1}^{x} n^r - (n-1)^r$$

have any of you seen this before? does it have a name? does it have an easy proof?

it gives an easy way to recursively define the formula for the sum I was looking for, which gave me this other interesting result that I'm sure is nothing new, but I just thought it was also cool:

$$\sum_{n=1}^{x} n^3 = [ \sum_{n=1}^{x} n]^2$$

2. Nov 4, 2004

### Gokul43201

Staff Emeritus
Yor first formula above : It's called a telescopic sum. Write out all the terms on the RHS and you'll find that they all cancel off, except the first and last terms, which gives you the LHS.

3. Nov 4, 2004

### StatusX

wow, thats simple, and completely different from the way i derived it, which was geometrical. still, its not completely trivial, because it does give the formulas i was after. for example:

$$x = \sum_{n=1}^{x} n - (n-1) = \sum_{n=1}^{x} 1 = x$$

$$x^2 = \sum_{n=1}^{x} n^2 - (n-1)^2 = \sum_{n=1}^{x} n^2-n^2 + 2n - 1 = 2 (\sum_{n=1}^{x} n) - x$$

which leaves:
$$\sum_{n=1}^{x} n = \frac{x(x+1)}{2}$$

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

which leaves:
$$\sum_{n=1}^{x} n^2 = \frac{x(x+1)(2x+1)}{6}$$

and so on.

Last edited: Nov 4, 2004
4. Nov 4, 2004

### Gokul43201

Staff Emeritus
Yes, that is a technique often used to find sums of powers. It's not trivial at all, when used for that.