Question Number 212432 by hardmath last updated on 13/Oct/24
$$\frac{\mathrm{9}}{\mathrm{2}\centerdot\mathrm{4}}\:+\:\frac{\mathrm{9}}{\mathrm{4}\centerdot\mathrm{6}}\:+…+\:\frac{\mathrm{9}}{\mathrm{4n}\centerdot\left(\mathrm{n}\:+\:\mathrm{1}\right)}\:=\:\frac{\mathrm{15}}{\mathrm{8}} \\ $$$$\mathrm{Find}:\:\:\boldsymbol{\mathrm{n}}\:=\:? \\ $$
Answered by Ar Brandon last updated on 13/Oct/24
$$\mathrm{9}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{4}{n}\left({n}+\mathrm{1}\right)}=\frac{\mathrm{15}}{\mathrm{8}} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{4}{n}}−\frac{\mathrm{1}}{\mathrm{4}\left({n}+\mathrm{1}\right)}\right)=\frac{\mathrm{5}}{\mathrm{24}} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{8}}\right)+\left(\frac{\mathrm{1}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{12}}\right)+\left(\frac{\mathrm{1}}{\mathrm{12}}−\frac{\mathrm{1}}{\mathrm{16}}\right)+ \\ $$$$\:\:\centerdot\centerdot\centerdot+\left(\frac{\mathrm{1}}{\mathrm{4}\left({n}−\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{4}{n}}\right)+\left(\frac{\mathrm{1}}{\mathrm{4}{n}}−\frac{\mathrm{1}}{\mathrm{4}\left({n}+\mathrm{1}\right)}\right)=\frac{\mathrm{5}}{\mathrm{24}} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}\left({n}+\mathrm{1}\right)}=\frac{\mathrm{5}}{\mathrm{24}}\:\Rightarrow\frac{\mathrm{1}}{\mathrm{4}\left({n}+\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{24}}\:\Rightarrow{n}+\mathrm{1}=\mathrm{6},\:{n}=\mathrm{5} \\ $$