Mathematical Proofs?

Question

Does anyone know the equation for solving 1^4 + 2^4 + 3^4 +....+ n ^4?

I know for 1+2+3+4+5+....n = n(n+1)/2
and 1^2 + 2^2 + 3^2 + ...+ n^2 = (n(n+1)(2n+1))/6
and 1^3 + 2^3 + 3^3 + ...+ n^3 = (n(n+1)/2)^2

please help me!!

Answer 1

IMO, a formula that is easier to work with is:

∑ x^4 = n^5/5 + n^4/2 + n^3/3 - n/30 ...for x = 1 to n


If you find the formulas for the summations of higher powers and write them in the above format, and you may be start to notice a pattern... sort of. For ∑ x^k, the first term is always x^(k+1)/(k+1), the second term is always x^k/2, and the third term is always kx^(k-1)/12. After that, you can predict subsequent terms if you know about Bernoulli numbers.

Answer 2

There's probably documentation elsewhere on the web, but, if I were going to try to solve this, I'd simply assume that the solution is a quintic with no constant term (since you'd like it to work for zero as well as 1 and higher) and plug in a few known values and the values for n-1 and n+1 and I'll bet that you can solve it.

So, let's assume that the solution is of the form:
f(n)=an^5+bn^4+cn^3+dn^2+en

we know that f(1)=1 and f(2)=17

so, a+b+c+d+d+e=1
32a+16b+8c+4d+2e=17

When I did this for the first 5 positive integers, I got a=1/5, b=1/2, c=1/3, d=0, and e=-1/30

So the fomula would be
(6n^5+15n^4+10n^3-n)/30

I checked it for the integers up to 57. I'm pretty sure that I have it right.

It would be fairly easy to prove by induction. I may give a supplemental answer that proves it.

Answer 3

Here is a proceedure for finding the answer:

Write
(k+1)^5 -k^5 =5k^4 +10k^3 +10k^2 +5k +1.

Now sum both sides from k=0 to k=n. The left hand side collapses to give
(1^5 -0^5)+(2^5-1^5)+...((n+1)^5-n^5 )=(n+1)^5

The right hand side gives
5*[sum of k^4]+10*[sum k^3]+
10*[sum k^2]+5*[sum k]+(n+1).

Now plug in the known results for the sums of k^3, k^2, and k. and solve for the sum of k^4.

Answer 4

n*(1+n)*(1+2*n)*(-1+3*n+3*n^2) / 30

First few values are: 1, 17, 98, 354, 979, 2275, 4676, 8772, 15333, 25333, 39974, 60710, 89271, 127687, 178312, 243848, 327369, 432345, 562666

Should be a straightforward proof by induction if you need it.

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