How can i find summation of the series $i^k$

Question

Series :

$$\sum_{i =1}^{n} i^k= 1^k+ 2^k + 3^k + 4^k +\ldots+n^k$$

where $k$ is a constant.

This does not seem to be Geometric progression , how can I evaluate the sum?

If possible if also want to find

$$\sum_{j =1}^{n} F(j)$$

where $F(j)$ is sum of the above series at $n=j$.

http://en.wikipedia.org/wiki/Faulhaber%27s_formula – kingW3 Jan 05 '15 at 13:06 — kingW3, Jan 05 '15 at 13:06
No, it is not @jinawee . – Thomas Andrews Jan 05 '15 at 13:14 — Thomas Andrews, Jan 05 '15 at 13:14

score 2 · Accepted Answer · answered Jan 05 '15 at 13:11

2

Take a look at Faulhaber's formula to see that

$$\sum_{i=1}^{n} i^k = \frac{1}{k+1}\sum_{j=1}^{k}(-1)^j\binom{k+1}{j}B_jn^{k+1-j}$$

where $B_j$ are the Bernoulli numbers.

answered Jan 05 '15 at 13:11

Aaron Maroja

see the edit , i also want to find its summation – Abhishek Gupta Jan 05 '15 at 13:16
2

@AbhishekGupta Well, you do the math. The formula is there. – Timbuc Jan 05 '15 at 13:22

score 2 · Answer 2 · answered Jan 05 '15 at 13:24

2

Alternately, look at the Bernoulli polynomials:
sums of the pth powers

answered Jan 05 '15 at 13:24

GEdgar

2 Answers2