Menu Close

1-3-2-3-3-3-4-3-x-3-1-4-2-7-3-10-x-3x-1-2021-find-x-




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

Leave a Reply

Your email address will not be published. Required fields are marked *