# Sum of the first n powers

1. Jul 24, 2007

### ehrenfest

1. The problem statement, all variables and given/known data
I encountered this formula in a proof:

Let $$S_k(n)$$ be the sum of the first n positive integers raised to the power of k:

e.g. $$S_3(5) = 1^3 + 2^3+3^3+4^3+5^3$$

How would one show that

$$S_2(n) = (n^3/3) + (n^2/2) + an$$

and that

$$S_4(n) = (n^5/5) + (n^4/2) +bn^3 + cn^2 +dn$$

where a,b,c,d are constants.

2. Relevant equations

3. The attempt at a solution

2. Jul 24, 2007

### rocomath

attempt? there needs to be at least some work b4 help is given.

3. Jul 24, 2007

### ehrenfest

3. The attempt at a solution

n^5/5 is the integral of n^4 and n^3/3 is the integral of n^2
However, that is only true for the first term.

4. Jul 24, 2007

### durt

When you're given a formula and asked to prove it, it's often helpful to try induction.

5. Jul 25, 2007

### Dick

You are guessing e.g. for S4 that it is a fifth degree polynomial. Then do something in the spirit of induction. What is S4(k)-S4(k-1)? Once you answer that you can solve for the coefficients.

6. Jul 25, 2007

### ehrenfest

S4(k)-S4(k-1) = k^4
So, I should set that equal to a(k)^5 + b(k)^4 + c(k)^3 +d(k)^2 + e(k) + f - [a(k-1)^5 + b(k-1)^4 + c(k-1)^3 +d(k-1)^2 + e(k-1) + f] ?
I am not sure how you solve that or how you would do induction on that.

7. Jul 25, 2007

### Dick

Expand everything out and move everything to one side. Now you've got a polynomial that must equal zero for all values of k. So the coefficients of all powers of k must be zero.