Sum of $k^4$ from $0$ to $n$

Question

How can I find this summation? I started by expanding $(k+1)^5$ and setting the summation of both equal to each other. There is some cancellation but I don't know what to do afterwards.

Answer 1

HINT:

$$(k+1)^5-k^5=\binom51k^4+\binom52k^3+\binom53k^2+\binom54k+1$$

Set $k=1,2\cdots,n$ and add to find $$(n+1)^5-1=5S_4+10S_3+10S_2+5S_1+S_0$$

where $S_m=\sum_{r=1}^nr^m$

More generally, $$(k+1)^{n+1}-k^{n+1}=\sum_{r=0}^n\binom{n+1}rS_r$$

Answer 2

One way that should be easy to remember is just linear algebra. it is guaranteed that the answer is $$ a_5 n^5 + a_4 n^4 + a_3 n^3 + a_2 n^2 + a_1 n + a_0, $$ where all six $a_j$ are rational numbers. They might all be positive, I'm not sure, but $a_5 > 0.$

So, the sum up to $n=1$ is $1,$ or $$ a_5 + a_4 + a_3 + a_2 + a_1 + a_0 = 1. $$

The sum up to $n=2$ is $17,$ or $$ 32 a_5 + 16 a_4 + 8 a_3 + 4 a_2 + 2 a_1 + a_0 = 17. $$

The sum up to $n=3$ is $98,$ or $$ 243 a_5 + 81 a_4 + 27 a_3 + 9 a_2 + 3 a_1 + a_0 = 98. $$

The sum up to $n=4$ is $354,$ or $$ 1024 a_5 + 256 a_4 + 64 a_3 + 16 a_2 + 4 a_1 + a_0 = 354. $$

The sum up to $n=5$ is $979,$ or $$ 3125 a_5 + 625 a_4 + 125 a_3 + 25 a_2 + 5 a_1 + a_0 = 979. $$

The sum up to $n=6$ is $2275,$ or $$ 7776 a_5 + 1296 a_4 + 216 a_3 + 36 a_2 + 6 a_1 + a_0 = 2275. $$

So, there you go, six linear equations in six unknowns. You just solve the system with Gauss-Jordan elimination. I saw one of your questions was on row echelon form, so you know how to do this.

There are also many clever ways to do this. Matter of taste.