Σ_(k = 1) ^n k^5 = ?


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Question Number 103796 by bemath last updated on 17/Jul/20

$\underset{k = 1}{\sum^{n}} k^{5} = ?$
Commented by mr W last updated on 17/Jul/20

Commented by bemath last updated on 17/Jul/20

$w a w . . c o m p l e t e s i r .$ $h a h a == . c o o l l$
	Answered by Ar Brandon last updated on 17/Jul/20

	$Consider;$ $k^{6} - {(k - 1)}^{6} = k^{6} - (k^{6} - {6 k}^{5} + {15 k}^{4} - {20 k}^{3} + {15 k}^{2} - 6 k + 1)$ $= {6 k}^{5} - {15 k}^{4} + {20 k}^{3} - {15 k}^{2} + 6 k - 1$ $\Rightarrow \underset{k = 1}{\sum^{n}} [k^{6} - {(k - 1)}^{6}] = 6 \underset{k = 1}{\sum^{n}} k^{5} + \underset{k = 1}{\sum^{n}} [- {15 k}^{4} + {20 k}^{3} - {15 k}^{2} + 6 k - 1]$ $\Rightarrow n^{6} = 6 \underset{k = 1}{\sum^{n}} k^{5} - 15 \underset{k = 1}{\sum^{n}} k^{4} + 20 \underset{k = 1}{\sum^{n}} k^{3} - 15 \underset{k = 1}{\sum^{n}} k^{2} + 6 \underset{k = 1}{\sum^{n}} k - \underset{k = 1}{\sum^{n}} (1)$ $S u b s t i t u t i n g t h e p r e v i o u s s e r i e s a n d m a k i n g \underset{k = 1}{\sum^{n}} k^{5}$ $t h e s u b j e c t, w e g e t o u r a n s w e r .$
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