# 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.

Aug 5, 2018

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

${N}^{9} / 9 + {N}^{8} / 2 + \frac{2 {N}^{7}}{3} - \frac{7 {N}^{5}}{15} + \frac{2 {N}^{3}}{9} - \frac{N}{30}$

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:

${B}_{n} = \frac{1}{6} , 0 , - \frac{1}{30} , 0 , \frac{1}{42} , 0 , - \frac{1}{30}$

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)

$= \frac{1}{12} \cdot 8 {N}^{7}$

$= \frac{2 {N}^{7}}{3}$

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

$= - \frac{1}{720} \cdot 336 {N}^{5}$

$= - \frac{7 {N}^{5}}{15}$

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

$= {N}^{9} / 9 + {N}^{8} / 2 + \frac{2 {N}^{7}}{3} - \frac{7 {N}^{5}}{15} + \frac{2 {N}^{3}}{9} - \frac{N}{30}$

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

and this gave:

Aug 5, 2018

$\text{Faulhaber's formula}$

#### Explanation:

$\text{Faulhaber's formula explains the how and why :}$

https://en.wikipedia.org/wiki/Faulhaber%27s_formula