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This question already has an answer here:

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 $n=j$.

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marked as duplicate by MJD, Hans Lundmark, Grigory M, Mark Bennet, user 170039 Jan 5 '15 at 13:40

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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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.

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  • $\begingroup$ see the edit , i also want to find its summation $\endgroup$ – Abhishek Gupta Jan 5 '15 at 13:16
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    $\begingroup$ @AbhishekGupta Well, you do the math. The formula is there. $\endgroup$ – Timbuc Jan 5 '15 at 13:22
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Alternately, look at the Bernoulli polynomials:
sums of the pth powers

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