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Question Number 2544 by Rasheed Soomro last updated on 22/Nov/15

C a n y o u G e n e r a l i z e t h e f o l l o w i n g ?

1 + 2 + 3 + \dots + n = \frac{1}{2} (n) (n + 1)

1^{2} + 2^{2} + 3^{2} + \dots + n^{2} = \frac{1}{6} (n) (n + 1) (2 n + 1

1^{3} + 2^{3} + 3^{3} + \dots + n^{3} = {[\frac{1}{2} (n) (n + 1)]}^{2}

\dots . \dots . . \dots . . \dots \dots . . \dots . . \dots \dots

\dots \dots \dots . \dots . \dots \dots

1^{k} + 2^{k} + 3^{k} + \dots + n^{k} = ? ? ?

Answered by Filup last updated on 23/Nov/15

\underset{i = 1}{\sum^{n}} i^{k} = H_{n}^{(- k)}

\underset{i = 0}{\sum^{\infty}} {(i + 1)}^{k} = ζ (- k) Re (k) + 1 < 0

\underset{i = 0}{\sum^{n}} {(i + 1)}^{k} = ζ (- k) - ζ (- k, n + 2)

Commented by Rasheed Soomro last updated on 22/Nov/15

I {d i d n}^{'} t u n d e r s t a n d!

W h a t a r e m, ζ (- k) a n d ζ (- k, m + 2) ?

F o r m u l a f o r 1^{k} + 2^{k} + \dots + n^{k} s h o u l d i n v o l v e o n l y

n a n d k . W h e r e d i d m c o m e f r o m ?

Commented by prakash jain last updated on 22/Nov/15

ζ is Riemann zeta function

ζ (z) = \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{z}}, ℜ (z) > 1

Commented by Filup last updated on 23/Nov/15

S o r r y, I did a direct copy from

W o l f r a m without adjusting for

your variables . I^{'} v e fixed it to suite

your question!

Commented by Rasheed Soomro last updated on 23/Nov/15

C ould you answer in simplified form . I me a n

without using riemann zeta notation ?