How would one systematically obtain the formula for $$ 1^k+2^k+3^k+\cdots+(n-1)^k+n^k$$ for a fixed positive integer exponent k? I am aware of the formulas $$\sum_{i=0}^n i^2 = \frac{(n^2+n)(2n+1)}{6}\quad ,\quad\sum_{i=0}^n i = \frac{n(n+1)}{2}\quad,\quad\sum_{i=0}^n i^3 = \left(\frac{n(n+1)}{2}\right)^2$$ i.e. when values of k are 1, 2 and 3. Using these can I use induction to proceed with the original sum?
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3You're looking for Faulhaber's Formula, see here for its derivation. – Workaholic Jul 04 '16 at 07:39
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Do you necessary want a proof using induction or are Ok with a different approach? – Olivier Oloa Jul 04 '16 at 07:44
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1@OlivierOloa I can go for a different approach, but i wanted to see if induction could handle it – Ayan Shah Jul 04 '16 at 07:49
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@Workaholic, Thankyou very much....I appreciate your help very much – Ayan Shah Jul 04 '16 at 07:51
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@AyanShah You're welcome! :-) – Workaholic Jul 04 '16 at 17:42
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A slightly different approach. One may use the Bernoulli polynomials satisfying $$B_{k+1}(x+1)-B_{k+1}(x)=(k+1)x^k$$ then, one gets $$\sum_{j=1}^n (k+1)j^k=\sum_{j=1}^n (B_{k+1}(j+1)-B_{k+1}(j))$$ but this is a telescoping sum giving
$$ \sum_{j=1}^n j^k=1^k+2^k+3^k+....+(n-1)^k+n^k=\frac{B_{k+1}(n+1)-B_{k+1}(1)}{k+1}. $$
Olivier Oloa
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