Question Number 144858 by mathdanisur last updated on 29/Jun/21
Answered by mindispower last updated on 29/Jun/21
$$\left({n}^{\mathrm{2}} +\mathrm{2}{n}+\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} =\left({n}+\mathrm{1}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} ={a}^{\mathrm{2}} \\ $$$$\left({n}^{\mathrm{2}} −\mathrm{2}{n}+\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} =\left(\left({n}−\mathrm{1}\right)^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} ={b}^{\mathrm{2}} \\ $$$$\left({n}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} =\boldsymbol{{ab}} \\ $$$$\frac{\mathrm{1}}{\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} +\boldsymbol{{ab}}}=\frac{\boldsymbol{{a}}−\boldsymbol{{b}}}{\boldsymbol{{a}}^{\mathrm{3}} −\boldsymbol{{b}}^{\mathrm{3}} }=\frac{\sqrt[{\mathrm{3}}]{\boldsymbol{{n}}+\mathrm{1}}−\sqrt[{\mathrm{3}}]{\boldsymbol{{n}}−\mathrm{1}}}{\mathrm{2}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\mathrm{999}} {\sum}}\frac{\sqrt[{\mathrm{3}}]{{n}+\mathrm{1}}−\sqrt[{\mathrm{3}}]{{n}−\mathrm{1}}}{\mathrm{2}}=\frac{\sqrt[{\mathrm{3}}]{\mathrm{1000}}+\sqrt[{\mathrm{3}}]{\mathrm{999}}−\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{9}+\sqrt[{\mathrm{3}}]{\mathrm{999}}}{\mathrm{2}} \\ $$$$ \\ $$
Commented by mathdanisur last updated on 29/Jun/21
$${alot}\:{perfect}\:{Sir}\:{thankyou} \\ $$
Commented by mathdanisur last updated on 30/Jun/21
$${Dear}\:{Sir},\:{but}\:{answer}\:\mathrm{5} \\ $$
Commented by mathdanisur last updated on 30/Jun/21
$${f}\left(\mathrm{1}\right)+{f}\left(\mathrm{2}\right)+{f}\left(\mathrm{3}\right)+...\:\boldsymbol{{no}},\:{f}\left(\mathrm{1}\right)+{f}\left(\mathrm{3}\right)+{f}\left(\mathrm{5}\right)+... \\ $$