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.

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Answer 1 · 2018-08-05T07:37:47

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 $n = 3, 5, 7$ vanish. The next nonzero one is $n = 4$ :

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

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

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

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

Once you work it out to $n = 8$ , you should get:

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

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

and this gave:

Answer 2 · 2018-08-05T08:02:44

$"Faulhaber's formula"$

$"Faulhaber's formula explains the how and why :"$