Power Sums: Coefficient Sum

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 $m=864$ , if $\displaystyle \sum_{i=1}^{864} a_i$ can be written as $\dfrac{p}{q}$ for coprime positive integers $p,q$ , find $p+q$ .

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The answer is 1729.

2 solutions

Trevor Arashiro
Feb 23, 2017

The sum of the coefficients of all power sums is 1. (hint, plug in n=1)

However, in this problem, we are missing $a_{865}$ . The lead coefficient of the closed form of the sum of mth powers is $\frac{1}{m+1}$ . Thus

$1-\frac{1}{865}=\frac{864}{865}\Longrightarrow 864+865=\boxed{1729}$

Apologies to anyone who answered 201. I thought I changed the problem to 864 but the only explanation I have as to why it didn't change is that I didn't hit save? I might have only hit preview.

Trevor Arashiro - 4 years, 3 months ago

i only induced that $a_{865}=\frac{1}{865}$ have u a more powerful proof?

Rohith M.Athreya - 4 years, 3 months ago

Suppose that some polynomial $h(x)$ exists such that $h(a)-h(a-1)=a^m$ . Then $\sum_{k=1}^n k^m=h(n)-h(0)$ . Next, assume $h(0)=0$ (you can actually assume h(0)=anything).

This means that $h(n)$ is the closed form of the polynomial solution that we desire. Looking back at $h(a)-h(a-1)=a^m$ , if we assume that $h(n)$ is a polynomial of degree m+1, we can write the first few terms as $\large{(}$ $a_{m+1}n^{m+1}+a_{m}n^m+......$ $\large{)}$ $-$ $\large{(}$ $a_{m+1}(n-1)^{m+1}+a_m(n-1)^{m}+....$ $\large{)}$ $=n^m$

If we look at the coefficients of $n^m$ , we get $a_{m+1}\binom{m+1}{1}n^m=n^m$ . It is quite clear that the lead coefficient must be $\frac{1}{m+1}$

Trevor Arashiro - 4 years, 3 months ago

Thank you!

Rohith M.Athreya - 4 years, 3 months ago

Chew-Seong Cheong
Feb 27, 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 . And that $\displaystyle \sum_{k=1}^{p+1} b_{pk} = 1$ ( ref: eqn (19) ).

For $p=m=864$ , we have $b_{864,i} = a_i$ and:

$\begin{aligned} \sum_{i=1}^{864} a_i & = \sum_{i=1}^{\color{#D61F06}865} a_i - a_{\color{#D61F06}865} \\ & = 1 - a_{\color{#D61F06}865} \\ & = 1 - b_{864,865} \\ & = 1 - \frac {(-1)^{864-865+1}B_{864-865+1}864!}{865!(864-865+1)!} \\ & = 1 - \frac {(-1)^0{\color{#3D99F6}B_0}864!}{865!0!} & \small \color{#3D99F6} \text{Note that } B_0 = 1 \\ & = 1 - \frac 1{865} \\ & = \frac {864}{865} \end{aligned}$

$\implies p+q = 864+865 = \boxed{1729}$

Power Sums: Coefficient Sum

The answer is 1729.

2 solutions

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