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abiy3000






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summation of powers of consecutive integers
posted on: 9/2/2006 5:37:22 AM

hi, is there a general formula for
(1^n+2^n+3^n+...+r^n)?
illoyd




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summation of powers of consecutive integers
replied on: 9/4/2006 3:22:31 PM

Sums of Powers

Let Sk[N] denote the sum of the kth powers of the first N integers.
Most people are familiar with the formulas for the first few sums

S1[N] = N(N+1)/2

S2[N] = N(N+1)(2N+1)/6

S3[N] = N^2 (N+1)^2 /4

S4[N] = N(N+1)(2N+1)(3N^2+3N-1)/30

etc.

The general formula can be expressed as

B_(k+1)[N+1] - B_(k+1)[0]
Sk[N] = --------------------------
k+1
illoyd




Quote | Reply |


summation of powers of consecutive integers
replied on: 9/4/2006 3:27:52 PM

Try again

The general formula can be expressed as

Sk[N] = (B_(k+1)[N+1] - B_(k+1)[0])/k+1

where B_n[x] denotes the nth Bernoulli polynomial. There are several
interesting methods for deriving these formulas. Around 1690 Jacques
Bernoulli gave a nice description of his derivation in a short paper
where he introduced the "Bernoulli Numbers".

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