Find a formula for (1^5 + 2^5 +3^5 +4^5 +………..+(n-1)^5 + n^5 =)?

Answer 1

http://polysum.tripod.com/

there...

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Answer 2

In general, I believe that the formula for 1^k + 2^k + ... + n^k for positive integral n, k can be represented by a polynomial of degree k+1. That being said,

1^5 + 2^5 + 3^5 + ... +n^5 = a + bn + cn^2 + dn^3 + en^4 + fn^5 + gn^6 where a, b, c, d, e, f, and g are coefficients of the polynomial. You could use this to generate a system of 7 eqns. with 7 unknowns and solve for the coefficients of that beastly thing. Of course, if you are not convinced that my conjecture about 1^k + 2^k + ...+ n^k and the polynomial of degree k+1 being true, then once you derive the 6th degree polynomial above you could use induction to prove they are equal.

Still this method is very computation heavy.

Answer 3

One method was created by Bernoulli. And actually coefficients that you've got in this expansions are C(n+1,k)*B(k), where B(k) - Bernoulli numbers.

His method worked like this:

(n+1)^6 - n^6 = C(6,5)n^5 + C(6,4)n^4 + C(6,3)n^3 + C(6,2)n^2 + C(6,1)n + 1
(n+1)^6 - 1 = ((n+1)^6 - n^6) + (n^6 - (n-1)^6) + ... + (2^6 - 1^6) =
= C(6,5)Sum(n^5) + C(6,4)Sum(n^4) + C(6,3)Sum(n^3) + C(6,2)Sum(n^2) + C(6,1)Sum(n) + n

Sum(n^5) = ((n+1)^6 - C(6,4)Sum(n^4) - C(6,3)Sum(n^3) - C(6,2)Sum(n^2) - C(6,1)Sum(n) - (n + 1))/C(6,5)

So you've got recursive formula. If you know expressions for the sums of all previous powers you can find the expression for next one.

My flavor of this approach (though may be I am not original here):
S(n) = a6*n^6 + a5*n^5 + a4*n^4 + a3*n^3 + a2*n^2 + a1*n + a0
S(n+1) - S(n) = a6((n+1)^6 - n^6) + a5*((n+1)^5 - n^5) + ... + a1 = n^5
If you expand this expression you've got triangular matrix for coefficients a6-a1 with right side (1, 0, 0, 0, 0, 0), which is very easily solvable even by hand.
a(0) = S(1) - (a6 + a5 + a4 + a3 + a2+ a1), where S(1) = 1.

Answer 4

n^5 ?

Answer 5

Arithmetic series formula,

Summation = 1/2 (n) (1^5+n^5)
n as the number of terms.
1^5 as the first term
and n^5 as the last term.

*meaning =1/2 times the number of terms times the sum of the first and the last term.