alternate form of sum of rth powers of the first n positive integers
We will show that
n∑k=0kr=∫n+11br(x)𝑑x |
We need two basic facts. First, a property of the Bernoulli polynomials is that b′r(x)=rbr-1(x). Second, the Bernoulli polynomials can be written as
br(x)=r∑k=1(rk)Br-kxk+Br |
We then have
∫n+11br(x) | =1r+1(br+1(n+1)-br+1(1))=1r+1r+1∑k=0(r+1k)Br+1-k((n+1)k-1) | ||
=1r+1r+1∑k=1(r+1k)Br+1-k(n+1)k |
Now reverse the order of summation (i.e. replace k by r+1-k) to get
∫n+11br(x)=1r+1r∑k=0(r+1r+1-k)Bk(n+1)r+1-k=1r+1r∑k=0(r+1k)Br(n+1)r+1-k |
which is equal to ∑nk=0kr (see the parent (http://planetmath.org/SumOfKthPowersOfTheFirstNPositiveIntegers) article).
Title | alternate form of sum of rth powers of the first n positive integers |
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Canonical name | AlternateFormOfSumOfRthPowersOfTheFirstNPositiveIntegers |
Date of creation | 2013-03-22 17:46:10 |
Last modified on | 2013-03-22 17:46:10 |
Owner | rm50 (10146) |
Last modified by | rm50 (10146) |
Numerical id | 4 |
Author | rm50 (10146) |
Entry type | Proof |
Classification | msc 11B68 |
Classification | msc 05A15 |