# Prove-that-1-3-2-3-3-3-n-3-1-2-3-n-2-

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