All power sums have a closed polynomial forms for integral powers. For example,

\[1^2+2^2+3^2+\cdots+n^2=\displaystyle \sum_{k=1}^n k^2=\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}\]

More generally

\[1^m+2^m+3^m+\cdots+n^m=\displaystyle \sum_{k=1}^n k^m=\displaystyle \sum_{i=1}^{m+1} a_i n^i\]

In the case of \(m=2\), \(a_1=\frac{1}{6}\), \(a_2=\frac{1}{2}\), and \(a_3=\frac{1}{3}\).

When \(m=2017\), find \(a_{2017}-a_{2015}\).

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