bomba923
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Quite Curious
\begin{gathered}<br /> \sum\limits_{i = 1}^n i = \frac{{n\left( {n + 1} \right)}}<br /> {2} \hfill \\<br /> \sum\limits_{i = 1}^n {i^2 } = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}<br /> {6} \hfill \\<br /> \sum\limits_{i = 1}^n {i^3 } = \frac{{n^2 \left( {n + 1} \right)^2 }}<br /> {4} \hfill \\<br /> \vdots \hfill \\<br /> \left( {etc} \right) \hfill \\ <br /> \end{gathered}
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But in general,
\forall k \in \mathbb{N} ,
what is the general summation formula for
\sum\limits_{i = 1}^n {i^k } \; {?}
\begin{gathered}<br /> \sum\limits_{i = 1}^n i = \frac{{n\left( {n + 1} \right)}}<br /> {2} \hfill \\<br /> \sum\limits_{i = 1}^n {i^2 } = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}<br /> {6} \hfill \\<br /> \sum\limits_{i = 1}^n {i^3 } = \frac{{n^2 \left( {n + 1} \right)^2 }}<br /> {4} \hfill \\<br /> \vdots \hfill \\<br /> \left( {etc} \right) \hfill \\ <br /> \end{gathered}
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But in general,
\forall k \in \mathbb{N} ,
what is the general summation formula for
\sum\limits_{i = 1}^n {i^k } \; {?}
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