Question Number 181260 by lapache last updated on 23/Nov/22
$${Calcul}\: \\ $$$$\underset{{n}=\mathrm{3}} {\overset{+\infty} {\sum}}\:\frac{\mathrm{2}{n}−\mathrm{1}}{{n}\left({n}+\mathrm{2}\right)\left({n}−\mathrm{2}\right)}=…?? \\ $$
Answered by SEKRET last updated on 23/Nov/22
$$\frac{\mathrm{89}}{\mathrm{96}} \\ $$
Answered by SEKRET last updated on 23/Nov/22
$$\sum_{\boldsymbol{\mathrm{n}}=\mathrm{3}} ^{\infty} \left(\:\frac{\mathrm{1}}{\mathrm{4}\boldsymbol{\mathrm{n}}}−\frac{\mathrm{5}}{\mathrm{8}\left(\boldsymbol{\mathrm{n}}+\mathrm{2}\right)}+\frac{\mathrm{3}}{\mathrm{8}\left(\boldsymbol{\mathrm{n}}−\mathrm{2}\right)}\right)= \\ $$$$=\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{3}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{4}\boldsymbol{\mathrm{n}}}\:−\:\frac{\mathrm{2}}{\mathrm{8}\left(\boldsymbol{\mathrm{n}}+\mathrm{2}\right)}\:−\:\frac{\mathrm{3}}{\mathrm{8}\left(\boldsymbol{\mathrm{n}}+\mathrm{2}\right)}\:+\:\frac{\mathrm{3}}{\mathrm{8}\left(\boldsymbol{\mathrm{n}}−\mathrm{2}\right)}\right)= \\ $$$$\:\frac{\mathrm{1}}{\mathrm{4}}\centerdot\underset{\boldsymbol{\mathrm{n}}=\mathrm{3}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}}\:−\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}+\mathrm{2}}\right)\:+\:\frac{\mathrm{3}}{\mathrm{8}}\underset{\boldsymbol{\mathrm{n}}=\mathrm{3}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}−\mathrm{2}}\:−\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}+\mathrm{2}}\right)=\boldsymbol{\mathrm{S}}_{\mathrm{1}} +\boldsymbol{\mathrm{S}}_{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{S}}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{4}}\centerdot\left(\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{8}}…\right)= \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}\centerdot\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}\right)=\frac{\mathrm{7}}{\mathrm{48}} \\ $$$$\boldsymbol{\mathrm{S}}_{\mathrm{2}} =\:\frac{\mathrm{3}}{\mathrm{8}}\centerdot\left(\frac{\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{9}}+\frac{\mathrm{1}}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{10}}+….\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{8}}\centerdot\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}\right)=\frac{\mathrm{3}}{\mathrm{8}}\centerdot\frac{\mathrm{12}+\mathrm{6}+\mathrm{4}+\mathrm{3}}{\mathrm{12}}=\frac{\mathrm{25}}{\mathrm{32}} \\ $$$$\boldsymbol{\mathrm{S}}_{\mathrm{1}} +\boldsymbol{\mathrm{S}}_{\mathrm{2}} =\frac{\mathrm{7}}{\mathrm{48}}+\frac{\mathrm{25}}{\mathrm{32}}=\frac{\mathrm{7}}{\mathrm{2}\centerdot\mathrm{3}\centerdot\mathrm{8}}+\frac{\mathrm{25}}{\mathrm{2}\centerdot\mathrm{2}\centerdot\mathrm{8}}=\frac{\mathrm{14}}{\mathrm{96}}+\frac{\mathrm{75}}{\mathrm{96}}=\frac{\mathrm{89}}{\mathrm{96}} \\ $$$$\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{3}} {\overset{\infty} {\sum}}\:\:\:\frac{\mathrm{2}\boldsymbol{\mathrm{n}}−\mathrm{1}}{\boldsymbol{\mathrm{n}}\:\left(\boldsymbol{\mathrm{n}}+\mathrm{2}\right)\left(\boldsymbol{\mathrm{n}}−\mathrm{2}\right)}\:\:=\:\:\frac{\mathrm{89}}{\mathrm{96}} \\ $$$$\boldsymbol{{ABDULAZIZ}}\:\:\:\boldsymbol{{ABDUVALIYEV}} \\ $$