how to think from 1+2+3+...+n=((n(n+1))/2) 1^2 +2^2 +3^2 +...+n^2 =((n(n+1)(2n+1))/6) 1^3 +2^3 +3^3 +...+n^3 =(((n(n+1))/2))^2


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Question Number 159330 by bounhome last updated on 15/Nov/21

$h o w t o t h i n k f r o m$ $1 + 2 + 3 + . . . + n = \frac{n (n + 1)}{2}$ $1^{2} + 2^{2} + 3^{2} + . . . + n^{2} = \frac{n (n + 1) (2 n + 1)}{6}$ $1^{3} + 2^{3} + 3^{3} + . . . + n^{3} = {(\frac{n (n + 1)}{2})}^{2}$
	Answered by Ar Brandon last updated on 15/Nov/21

	$S = 1 + 2 + 3 + . . . (n - 2) + (n - 1) + n$ $S = n + (n - 1) + (n - 2) . . . + 3 + 2 + 1$ $2 S = (n + 1) + (n + 1) + (n + 1) + . . . + (n + 1) + (n + 1) + (n + 1)$ $= n (n + 1) \Rightarrow S = \frac{n (n + 1)}{2}$
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