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Question Number 65587 by naka3546 last updated on 31/Jul/19
Prove that 1^3 + 2^3 + 3^3 + … + n^3 = (1+2+3+...+n)^2
$${Prove}\:\:{that} \\ $$$$\:\:\:\:\:\mathrm{1}^{\mathrm{3}} \:+\:\mathrm{2}^{\mathrm{3}} \:+\:\mathrm{3}^{\mathrm{3}} \:+\:\ldots\:+\:{n}^{\mathrm{3}} \:\:=\:\:\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+…+{n}\right)^{\mathrm{2}} \\ $$
Commented by naka3546 last updated on 31/Jul/19
No using Mathematical Induction .
$${No}\:\:{using}\:\:{Mathematical}\:\:{Induction}\:. \\ $$
Commented by Tanmay chaudhury last updated on 31/Jul/19
Commented by Tanmay chaudhury last updated on 31/Jul/19
S_3 =3∫S_2 dn+n×B_3 =3∫((n(n+1)(2n+1))/6)dn+n×0 =(1/2)∫n(2n^2 +3n+1)dn =(1/2)∫(2n^3 +3n^2 +n) dn (1/2)(((2n^4 )/4)+((3n^3 )/3)+(n^2 /2)) (1/2)(((6n^4 +12n^3 +6n^2 )/(12))) (1/4)(n^4 +2n^3 +n^2 ) (n^2 /4)(n^2 +2n+1) {((n(n+1))/2)}^2 Answer
$${S}_{\mathrm{3}} =\mathrm{3}\int{S}_{\mathrm{2}} {dn}+{n}×{B}_{\mathrm{3}} \\ $$$$=\mathrm{3}\int\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}{dn}+{n}×\mathrm{0} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int{n}\left(\mathrm{2}{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{1}\right){dn} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{2}{n}^{\mathrm{3}} +\mathrm{3}{n}^{\mathrm{2}} +{n}\right)\:\:{dn} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{2}{n}^{\mathrm{4}} }{\mathrm{4}}+\frac{\mathrm{3}{n}^{\mathrm{3}} }{\mathrm{3}}+\frac{{n}^{\mathrm{2}} }{\mathrm{2}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{6}{n}^{\mathrm{4}} +\mathrm{12}{n}^{\mathrm{3}} +\mathrm{6}{n}^{\mathrm{2}} }{\mathrm{12}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left({n}^{\mathrm{4}} +\mathrm{2}{n}^{\mathrm{3}} +{n}^{\mathrm{2}} \right) \\ $$$$\frac{{n}^{\mathrm{2}} }{\mathrm{4}}\left({n}^{\mathrm{2}} +\mathrm{2}{n}+\mathrm{1}\right) \\ $$$$\left\{\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\right\}^{\mathrm{2}} {Answer} \\ $$
Commented by Tanmay chaudhury last updated on 31/Jul/19
Answered by som(math1967) last updated on 31/Jul/19
let s=1^3 +2^3 +3^3 +....+n^3 now n^4 −(n−1)^4 =4n^3 −6n^2 +4n−1 ∴1^4 −0=4.1^3 −6.1^2 +4.1 −1 2^4 −1^4 =4.2^3 −6.2^2 +4.2−1 ......... n^4 −(n−1)^4 =4n^3 −6n^2 +4n−1 add n^4 =4.(1^3 +2^3 +.....n^3 )−6(1^2 +2^2 +...n^2 ) +4(1+2+..n)−1×n n^4 =4s−6×((n(n+1)(2n+1))/6) +4×((n(n+1))/2)−n 4s=n^4 +n(n+1)(2n+1) −2n(n+1)+n 4s=n(n^3 +1) +n(n+1)(2n+1−2) 4s=n(n+1)(n^2 −n+1)+n(n+1)(2n−1) 4s=n(n+1)(n^2 −n+1+2n−1) 4s=n(n+1)n(n+1) ∴s={((n(n+1))/2)}^2 ∴s=(1+2+3+.......n)^2
$${let}\:{s}=\mathrm{1}^{\mathrm{3}} +\mathrm{2}^{\mathrm{3}} +\mathrm{3}^{\mathrm{3}} +….+{n}^{\mathrm{3}} \\ $$$${now}\:{n}^{\mathrm{4}} −\left({n}−\mathrm{1}\right)^{\mathrm{4}} =\mathrm{4}{n}^{\mathrm{3}} −\mathrm{6}{n}^{\mathrm{2}} +\mathrm{4}{n}−\mathrm{1} \\ $$$$\therefore\mathrm{1}^{\mathrm{4}} −\mathrm{0}=\mathrm{4}.\mathrm{1}^{\mathrm{3}} −\mathrm{6}.\mathrm{1}^{\mathrm{2}} +\mathrm{4}.\mathrm{1}\:−\mathrm{1} \\ $$$$\mathrm{2}^{\mathrm{4}} −\mathrm{1}^{\mathrm{4}} =\mathrm{4}.\mathrm{2}^{\mathrm{3}} −\mathrm{6}.\mathrm{2}^{\mathrm{2}} +\mathrm{4}.\mathrm{2}−\mathrm{1} \\ $$$$……… \\ $$$${n}^{\mathrm{4}} −\left({n}−\mathrm{1}\right)^{\mathrm{4}} =\mathrm{4}{n}^{\mathrm{3}} −\mathrm{6}{n}^{\mathrm{2}} +\mathrm{4}{n}−\mathrm{1} \\ $$$${add} \\ $$$${n}^{\mathrm{4}} =\mathrm{4}.\left(\mathrm{1}^{\mathrm{3}} +\mathrm{2}^{\mathrm{3}} +…..{n}^{\mathrm{3}} \right)−\mathrm{6}\left(\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +…{n}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{4}\left(\mathrm{1}+\mathrm{2}+..{n}\right)−\mathrm{1}×{n} \\ $$$${n}^{\mathrm{4}} =\mathrm{4}{s}−\mathrm{6}×\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}\:+\mathrm{4}×\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}−{n} \\ $$$$\mathrm{4}{s}={n}^{\mathrm{4}} +{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)\:−\mathrm{2}{n}\left({n}+\mathrm{1}\right)+{n} \\ $$$$\mathrm{4}{s}={n}\left({n}^{\mathrm{3}} +\mathrm{1}\right)\:+{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}−\mathrm{2}\right) \\ $$$$\mathrm{4}{s}={n}\left({n}+\mathrm{1}\right)\left({n}^{\mathrm{2}} −{n}+\mathrm{1}\right)+{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}−\mathrm{1}\right) \\ $$$$\mathrm{4}{s}={n}\left({n}+\mathrm{1}\right)\left({n}^{\mathrm{2}} −{n}+\mathrm{1}+\mathrm{2}{n}−\mathrm{1}\right) \\ $$$$\mathrm{4}{s}={n}\left({n}+\mathrm{1}\right){n}\left({n}+\mathrm{1}\right) \\ $$$$ \\ $$$$\therefore{s}=\left\{\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\right\}^{\mathrm{2}} \\ $$$$\therefore{s}=\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+…….{n}\right)^{\mathrm{2}} \\ $$

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