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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

$(k+1)}^{p+1}-{k}^{p+1}=\sum _{s=0}^{p}\left(\begin{array}{c}p+1\\ s\end{array}\right){k}^{s$

hence we can find

$\sum _{k=1}^{n}{k}^{p}=\frac{1}{p+1}({(n+1)}^{p+1}-\sum _{s=0}^{p-1}\left(\begin{array}{c}p+1\\ s\end{array}\right)\sum _{k=1}^{n}{k}^{s}-1)$

then since by telescoping

and

hence we can find

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