# alternate form of sum of $r$th powers of the first $n$ positive integers

We will show that

 $\sum_{k=0}^{n}k^{r}=\int_{1}^{n+1}b_{r}(x)dx$

We need two basic facts. First, a property of the Bernoulli polynomials   is that $b_{r}^{\prime}(x)=rb_{r-1}(x)$. Second, the Bernoulli polynomials can be written as

 $b_{r}(x)=\sum_{k=1}^{r}\binom{r}{k}B_{r-k}x^{k}+B_{r}$

We then have

 $\displaystyle\int_{1}^{n+1}b_{r}(x)$ $\displaystyle=\frac{1}{r+1}(b_{r+1}(n+1)-b_{r+1}(1))=\frac{1}{r+1}\sum_{k=0}^{% r+1}\binom{r+1}{k}B_{r+1-k}((n+1)^{k}-1)$ $\displaystyle=\frac{1}{r+1}\sum_{k=1}^{r+1}\binom{r+1}{k}B_{r+1-k}(n+1)^{k}$

Now reverse the order of summation (i.e. replace $k$ by $r+1-k$) to get

 $\int_{1}^{n+1}b_{r}(x)=\frac{1}{r+1}\sum_{k=0}^{r}\binom{r+1}{r+1-k}B_{k}(n+1)% ^{r+1-k}=\frac{1}{r+1}\sum_{k=0}^{r}\binom{r+1}{k}B_{r}(n+1)^{r+1-k}$

which is equal to $\sum_{k=0}^{n}k^{r}$ (see the parent (http://planetmath.org/SumOfKthPowersOfTheFirstNPositiveIntegers) article).

Title alternate form of sum of $r$th powers of the first $n$ positive integers AlternateFormOfSumOfRthPowersOfTheFirstNPositiveIntegers 2013-03-22 17:46:10 2013-03-22 17:46:10 rm50 (10146) rm50 (10146) 4 rm50 (10146) Proof msc 11B68 msc 05A15