Have You Discovered This Recursive Formula for Sums of Powers?

[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

 

[Gokul43201]

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.

 

[StatusX]

wow, that's 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.

 

[Gokul43201]

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