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

We will show that

$$\sum _{k=0}^{n}{k}^{r}={\int}_{1}^{n+1}{b}_{r}(x)\mathit{d}x$$ |

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

$${b}_{r}(x)=\sum _{k=1}^{r}\left(\genfrac{}{}{0pt}{}{r}{k}\right){B}_{r-k}{x}^{k}+{B}_{r}$$ |

We then have

${\int}_{1}^{n+1}}{b}_{r}(x)$ | $={\displaystyle \frac{1}{r+1}}({b}_{r+1}(n+1)-{b}_{r+1}(1))={\displaystyle \frac{1}{r+1}}{\displaystyle \sum _{k=0}^{r+1}}\left({\displaystyle \genfrac{}{}{0pt}{}{r+1}{k}}\right){B}_{r+1-k}({(n+1)}^{k}-1)$ | ||

$={\displaystyle \frac{1}{r+1}}{\displaystyle \sum _{k=1}^{r+1}}\left({\displaystyle \genfrac{}{}{0pt}{}{r+1}{k}}\right){B}_{r+1-k}{(n+1)}^{k}$ |

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

$${\int}_{1}^{n+1}{b}_{r}(x)=\frac{1}{r+1}\sum _{k=0}^{r}\left(\genfrac{}{}{0pt}{}{r+1}{r+1-k}\right){B}_{k}{(n+1)}^{r+1-k}=\frac{1}{r+1}\sum _{k=0}^{r}\left(\genfrac{}{}{0pt}{}{r+1}{k}\right){B}_{r}{(n+1)}^{r+1-k}$$ |

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 |
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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 |