What is this formula I found?

  • Thread starter StatusX
  • Start date
  • #1
StatusX
Homework Helper
2,564
1
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:

[tex] x^r = \sum_{n=1}^{x} n^r - (n-1)^r [/tex]

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:

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

Answers and Replies

  • #2
Gokul43201
Staff Emeritus
Science Advisor
Gold Member
7,051
18
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
StatusX
Homework Helper
2,564
1
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:

[tex] x = \sum_{n=1}^{x} n - (n-1) = \sum_{n=1}^{x} 1 = x[/tex]

[tex] 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 [/tex]

which leaves:
[tex] \sum_{n=1}^{x} n = \frac{x(x+1)}{2}[/tex]

[tex] 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 [/tex]

which leaves:
[tex] \sum_{n=1}^{x} n^2 = \frac{x(x+1)(2x+1)}{6}[/tex]

and so on.
 
Last edited:
  • #4
Gokul43201
Staff Emeritus
Science Advisor
Gold Member
7,051
18
Yes, that is a technique often used to find sums of powers. It's not trivial at all, when used for that.
 

Related Threads on What is this formula I found?

Replies
6
Views
3K
  • Last Post
Replies
2
Views
653
  • Last Post
Replies
2
Views
2K
Replies
4
Views
1K
Replies
2
Views
2K
Replies
4
Views
1K
  • Last Post
Replies
21
Views
7K
Replies
7
Views
5K
Replies
15
Views
3K
Replies
4
Views
3K
Top