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

Σ_(k=1) ^n k^3 =[((n(n+1) )/2)]^2

$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{3}} =\left[\frac{{n}\left({n}+\mathrm{1}\right)\:}{\mathrm{2}}\right]^{\mathrm{2}} \\ $$

Commented by aliesam last updated on 04/Dec/19

god bless you

$${god}\:{bless}\:{you} \\ $$

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.

$${let}\:{P}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}}{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} \:\:\:{we}\:{have} \\ $$$${P}\left({x}+\mathrm{1}\right)−{P}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left({x}+\mathrm{2}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$=\frac{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}\left\{\:\:\left({x}+\mathrm{2}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} \right\}\:=\frac{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}\left\{\:{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{4}−{x}^{\mathrm{2}} \right\} \\ $$$$=\left({x}+\mathrm{1}\right)^{\mathrm{3}} \:\Rightarrow{P}\left({x}\right)−{P}\left({x}−\mathrm{1}\right)\:={x}^{\mathrm{3}} \:\Rightarrow\sum_{{k}=\mathrm{1}} ^{{n}} \:{k}^{\mathrm{3}} =\sum_{{k}=\mathrm{1}} ^{{n}} \left({P}\left({k}\right)−{P}\left({k}−\mathrm{1}\right)\right) \\ $$$$={P}\left(\mathrm{1}\right)−{P}\left(\mathrm{0}\right)+{P}\left(\mathrm{2}\right)−{P}\left(\mathrm{1}\right)+....+{P}\left({n}\right)−{P}\left({n}−\mathrm{1}\right) \\ $$$$={P}\left({n}\right)−{P}\left(\mathrm{0}\right)=\frac{\mathrm{1}}{\mathrm{4}}{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{0}\:=\frac{\left({n}\left({n}+\mathrm{1}\right)\right)^{\mathrm{2}} }{\mathrm{4}}\:\:{so}\:{the}\:{equaly}\:{is} \\ $$$${proved}. \\ $$

Commented by mathmax by abdo last updated on 04/Dec/19

you are welcome.

$${you}\:{are}\:{welcome}. \\ $$

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