Power Sums: Large Coefficients

Calculus Level 5

All power sums have a closed polynomial forms for integral powers. For example,

$1^2+2^2+3^2+\cdots+n^2=\displaystyle \sum_{k=1}^n k^2=\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}$

More generally

$1^m+2^m+3^m+\cdots+n^m=\displaystyle \sum_{k=1}^n k^m=\displaystyle \sum_{i=1}^{m+1} a_i n^i$

In the case of $m=2$ , $a_1=\frac{1}{6}$ , $a_2=\frac{1}{2}$ , and $a_3=\frac{1}{3}$ .

When $m=20$ , if $a_5$ can be written as $\dfrac{-p}{q}$ for coprime positive integers $p,q$ , find $p+q$ .

Tip: This might help :)

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The answer is 68733.

2 solutions

Chew-Seong Cheong
Feb 28, 2017

A power sum is given by $\displaystyle \sum_{k=1}^n k^p = \sum_{k=1}^{p+1} b_{pk} n^k$ ( ref: eqn (17) ), whose coefficient $b_{pk} = \dfrac {(-1)^{p-k+1}B_{p-k+1}p!}{k!(p-k+1)!}$ ( ref: eqn (18) ), where $B_m$ is a Bernoulli number .

For $p=m=20$ , we have $b_{20,i} = a_i$ and:

$\begin{aligned} a_5 & = b_{20,5} \\ & = \frac {(-1)^{20-5+1}B_{20-5+1}20!}{5!(20-5+1)!} \\ & = \frac {(-1)^{16}{\color{#3D99F6}B_{16}}20!}{5!16!} & \small \color{#3D99F6} \text{Note that }B_{16} = - \frac {3617}{510}\\ & = - \frac {969 \cdot 3617}{510} \\ & = - \frac {68723}{10} \end{aligned}$

$\implies p + q = 68723 + 10 = \boxed{68733}$

Trevor Arashiro
Feb 23, 2017

Take the bottom left half of Pascal's infinite matrix (linked in the problem to a google spread sheet). Remove the top left-bottom right diagonal of 1's. Remove the top row and right most column of 0's. Invert the matrix. The jth row from the right yields the coefficients of the sum of (j-1)th powers. This is proved in this document .

Power Sums: Large Coefficients

The answer is 68733.

2 solutions

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