Question Number 150769 by mathdanisur last updated on 15/Aug/21
$$\frac{\mathrm{1}^{\mathrm{3}} +\mathrm{2}^{\mathrm{3}} +\mathrm{3}^{\mathrm{3}} +\mathrm{4}^{\mathrm{3}} +...+\boldsymbol{\mathrm{x}}^{\mathrm{3}} }{\mathrm{1}\centerdot\mathrm{4}+\mathrm{2}\centerdot\mathrm{7}+\mathrm{3}\centerdot\mathrm{10}+...+\boldsymbol{\mathrm{x}}\left(\mathrm{3}\boldsymbol{\mathrm{x}}+\mathrm{1}\right)}\:=\:\mathrm{2021} \\ $$$$\mathrm{find}\:\:\boldsymbol{\mathrm{x}}=? \\ $$
Answered by nimnim last updated on 15/Aug/21
![Let me give a try.. (([((x(x+1))/2)]^2 )/(3×((x(x+1)(2x+1))/6)+((x(x+1))/2)))=2021 ⇒(((x(x+1))/2)/(2x+1+1))=2021 ⇒(x/4)=2021 ⇒ x=8084](Q150771.png)
$${Let}\:{me}\:{give}\:{a}\:{try}.. \\ $$$$\:\:\:\frac{\left[\frac{{x}\left({x}+\mathrm{1}\right)}{\mathrm{2}}\right]^{\mathrm{2}} }{\mathrm{3}×\frac{{x}\left({x}+\mathrm{1}\right)\left(\mathrm{2}{x}+\mathrm{1}\right)}{\mathrm{6}}+\frac{{x}\left({x}+\mathrm{1}\right)}{\mathrm{2}}}=\mathrm{2021} \\ $$$$\Rightarrow\frac{\frac{{x}\left({x}+\mathrm{1}\right)}{\mathrm{2}}}{\mathrm{2}{x}+\mathrm{1}+\mathrm{1}}=\mathrm{2021} \\ $$$$\Rightarrow\frac{{x}}{\mathrm{4}}=\mathrm{2021} \\ $$$$\Rightarrow\:{x}=\mathrm{8084} \\ $$
Commented by mathdanisur last updated on 15/Aug/21
$$\mathrm{Thankyou}\:\mathrm{Ser} \\ $$