calculate-k-1-n-k-4-interms-of-n-

Question Number 38640 by maxmathsup by imad last updated on 27/Jun/18

c a l c u l a t e \sum_{k = 1}^{n} k^{4} i n t e r m s o f n .

Commented by math khazana by abdo last updated on 01/Jul/18

w e h a v e {(k + 1)}^{5} - k^{5}

= \sum_{p = 0}^{5} C_{5}^{p} k^{p} - k^{5}

1 + C_{5}^{1} k + C_{5}^{2} k^{2} + C_{5}^{3} k^{3} + C_{5}^{4} k^{4} + C_{5}^{5} k^{5} - k^{5}

= 1 + 5 k + 10 k^{2} + 10 k^{3} + 5 k^{4} \Rightarrow

\sum_{k = 1}^{n} {{(k + 1)}^{5} - k^{5}}

= n + 5 \sum_{k = 1}^{n} k + 10 \sum_{k = 1}^{n} k^{2} + 10 \sum_{k = 1}^{n} k^{3} + 5 \sum_{k = 1}^{n} k^{4}

{(n + 1)}^{5} = n + 1 + 5 \frac{n (n + 1)}{2} + 10 \frac{n (n + 1) (2 n + 1)}{6}

+ 10 \frac{n^{2} {(n + 1)}^{2}}{4} + 5 \sum_{k = 1}^{n} k^{4} a f t e r c a l c u l u s w e f i n d

\sum_{k = 1}^{n} k^{4} = \frac{n^{5}}{5} + \frac{n^{4}}{2} + \frac{n^{3}}{3} - \frac{n}{30} .