# Is there any easy to solve this sigma question Find the sum of 1^8 +2^8 +3^8 +4^8+......to n terms? I tried by taking (n+1)^9 -(n) ^9 but that method is too long. Please anybody suggest easier and short method.

##### 2 Answers

Well, no easy way, no... It is apparently:

#N^9/9 + N^8/2 + (2 N^7)/3 - (7 N^5)/15 + (2 N^3)/9 - N/30#

It really isn't obvious, but as suggested, Faulhaber has found a solution that translates into this result:

#sum_(n=1)^(N) n^8 = N^9/9 + 1/2 N^8 + sum_(n=2)^(8) [B_n/(n!) (8!)/((8-n+1)!)]N^(8-n+1)# where

#B_n# is a Bernoulli number.

The relevant Bernoulli numbers are:

#B_n = 1/6, 0, -1/30, 0, 1/42, 0, -1/30# for

#n = 2, 3, 4, 5, 6, 7, 8# , respectively.

For example, the next term in the resultant formula is:

#B_2/(2!) (8!)/((8 - 2 + 1)!)N^(8-2+1)#

#= (1//6)/2 (7! cdot 8)/(7!)N^(7)#

#= 1/12 cdot 8N^7#

#= (2N^7)/3#

The terms of

#B_4/(4!) (8!)/((8 - 4 + 1)!)N^(8-4+1)#

#= (-1//30)/(24) (5! cdot 6 cdot 7 cdot 8)/(5!) N^5#

#= -1/(720) cdot 336 N^5#

#= -(7N^5)/15#

Once you work it out to

#= N^9/9 + N^8/2 + (2 N^7)/3 - (7 N^5)/15 + (2 N^3)/9 - N/30#

What I would have done (which is arguably easier) is just put it into Excel.

and this gave: