Σ_(k=1) ^n k^3 =[((n(n+1) )/2)]^2


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Question Number 74923 by aliesam last updated on 04/Dec/19

$\underset{k = 1}{\sum^{n}} k^{3} = {[\frac{n (n + 1)}{2}]}^{2}$
Commented by aliesam last updated on 04/Dec/19

$g o d b l e s s y o u$
Commented by mathmax by abdo last updated on 04/Dec/19
$let P(x)=(1/4)x^2 (x+1)^2 we have P(x+1)−P(x)=(1/4)(x+1)^2 (x+2)^2 −(1/4)x^2 (x+1)^2 =(((x+1)^2 )/4){ (x+2)^2 −x^2 } =(((x+1)^2 )/4){ x^2 +4x+4−x^2 } =(x+1)^3 ⇒P(x)−P(x−1) =x^3 ⇒Σ_(k=1) ^n k^3 =Σ_(k=1) ^n (P(k)−P(k−1)) =P(1)−P(0)+P(2)−P(1)+....+P(n)−P(n−1) =P(n)−P(0)=(1/4)n^2 (n+1)^2 −0 =(((n(n+1))^2 )/4) so the equaly is proved.$
$l e t P (x) = \frac{1}{4} x^{2} {(x + 1)}^{2} w e h a v e$ $P (x + 1) - P (x) = \frac{1}{4} {(x + 1)}^{2} {(x + 2)}^{2} - \frac{1}{4} x^{2} {(x + 1)}^{2}$ $= \frac{{(x + 1)}^{2}}{4} {{(x + 2)}^{2} - x^{2}} = \frac{{(x + 1)}^{2}}{4} {x^{2} + 4 x + 4 - x^{2}}$ $= {(x + 1)}^{3} \Rightarrow P (x) - P (x - 1) = x^{3} \Rightarrow \sum_{k = 1}^{n} k^{3} = \sum_{k = 1}^{n} (P (k) - P (k - 1))$ $= P (1) - P (0) + P (2) - P (1) + . . . . + P (n) - P (n - 1)$ $= P (n) - P (0) = \frac{1}{4} n^{2} {(n + 1)}^{2} - 0 = \frac{{(n (n + 1))}^{2}}{4} s o t h e e q u a l y i s$ $p r o v e d .$
Commented by mathmax by abdo last updated on 04/Dec/19

$y o u a r e w e l c o m e .$
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