Find out the answer to "Proof for ∑k=1nka equaling a sum of fractions"

nrgiizr0ib6

Answered question

2022-04-28

Proof for

\sum_{k = 1}^{n} k^{a}

equaling a sum of fractions

zavesiljid

Beginner2022-04-29Added 21 answers

Assuming that we know the expression of

\sum_{k = 1}^{n} k, \sum_{k = 1}^{n} k^{2}, \dots, \sum_{k = 1}^{n} k^{p - 1}

then since by telescoping

\sum_{k = 1}^{n} {(k + 1)}^{p + 1} - k^{p + 1} = {(n + 1)}^{p + 1} - 1

and

{(k + 1)}^{p + 1} - k^{p + 1} = \sum_{s = 0}^{p} (\begin{matrix} p + 1 \\ s \end{matrix}) k^{s}

hence we can find

\sum_{k = 1}^{n} k^{p} = \frac{1}{p + 1} ({(n + 1)}^{p + 1} - \sum_{s = 0}^{p - 1} (\begin{matrix} p + 1 \\ s \end{matrix}) \sum_{k = 1}^{n} k^{s} - 1)

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?