Question

First of all, sorry to post a similar question to the one I asked before. I do not know if that is allowed, my apologies.

Calculate the sum from k=1 to infinity (or n) of
(k^4) - (k-4)^4

The book says that the answer is n^4, but I find this incoherent as this is the answer for (k^4) - (k-1)^4

I obtain that the sum equals to
-f(-3) - f(-2) - f(-1) -f(0) + f(n-3) + f(n-2) + f(n-1) + f(n)

I assumed that f(k) = k^4 and f(k-1) = (k-1)^4
I would appreciate any correction or suggestion thanks ^-^

Answer 1

I am going to expand the expression, then use the formulas for ∑ n^3, ∑ n^2, etc

∑ (k^4) - (k-4)^4 for k = 1 to n
= ∑ k^4 - (k^4 - 16k^3 + 96k^2 -256k + 256) for k = 1 to n
=∑ 16k^3 - 96k^2 + 256k - 256 for k = 1 to n
= 16( n(n+1)/2 )^2 - 96n(n+1)(2n+1)/6 + 256n(n+1)/2 - 256n

this reduced to <b>4n(n^3 - 6n^2 + 21n - 36) or 4n^4 - 24n^3 + 84n^2 - 144n </b> , as confirmed by Wolfram:
http://www.wolframalpha.com/input/?i=expand+16(+n(n%2B1)%2F2+)%5E2+-+96n(n%2B1)(2n%2B1)%2F6+%2B+256n(n%2B1)%2F2+-+256n

using the original expression in Wolfram gave me the same answer
http://www.wolframalpha.com/input/?i=%E2%88%91+(k%5E4+-+(k-4)%5E4)+for+k+%3D+1+to+n