## Sum of First n m-th Powers

This is going to be long even without the proof.

The sum of the first n m-th powers is given by the formula :

where S2 is Stirling numbers of the Second Kind :

m\k| 1 2 3 4 5 6

——————————-

1 | 1

2 | 1 1

3 | 1 3 1

4 | 1 7 6 1

5 | 1 15 25 10 1

5 | 1 31 90 65 15 1

given by the recurrence relation :

for m >= 1,

S2(m,0) = 0,

S2(m,1) = 1,

S2(m,k) = 0 for all all k > m,

**S2(m+1,k) = S2(m,k-1) + k*S2(m,k).**

and **f(x,k) = x*(x-1)*(x-2)*…*(x-k+1).**

Let us take two examples :

T(2,n) = S2(2,1)*f(n+1,2)/2 + S2(2,2)*f(n+1,3)/3,

= 1*(n+1)*n/2 + 1*(n+1)*n*(n-1)/3,

= n*(n+1)*(1/2 + [n-1]/3),

= n*(n+1)*(1/2 + n/3 – 1/3),

= n*(n+1)*(2*n+1)/6.

and

T(4,n) = S2(4,1)*f(n+1,2)/2 + S2(4,2)*f(n+1,3)/3 +

S2(4,3)*f(n+1,4)/4 + S2(4,4)*f(n+1,5)/5,

= 1*(n+1)*n/2 + 7*(n+1)*n*(n-1)/3 +

6*(n+1)*n*(n-1)*(n-2)/4 +

1*(n+1)*n*(n-1)*(n-2)*(n-3)/5,

= n*(n+1)*(1/2 + 7*[n-1]/3 + 3*[n-1]*[n-2]/2 +

[n-1]*[n-2]*[n-3]/5),

= n*(n+1)*(1/2 + 7*[n-1]/3 + 3*[n^2-3*n+2]/2 +

[n^3-6*n^2+11*n-6]/5),

= n*(n+1)*(1/2 + 7*n/3 – 7/3 + 3*n^2/2 – 9*n/2 + 3 +

n^3/5 – 6*n^2/5 + 11*n/5 – 6/5),

= n*(n+1)*(n^3/5 + 3*n^2/10 + n/30 – 1/30),

= n*(n+1)*(6*n^3+9*n^2+n-1)/30,

= n*(n+1)*(2*n+1)*(3*n^2+3*n-1)/30.

Try higher powers for yourself.

-fR0D

Hello. Could you tell me where I can find a full proof of this formula?

Thanks,

Howard

Howard WalkerSeptember 13, 2012 at 2:49 AM

See problem solving strategies, chapter 5-Enumerative combinatorics

MasumOctober 20, 2012 at 4:40 PM