9-2-4-9-4-6-9-4n-n-1-15-8-Find-n-

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

$$\mathrm{9}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{4}{n}\left({n}+\mathrm{1}\right)}=\frac{\mathrm{15}}{\mathrm{8}} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{4}{n}}−\frac{\mathrm{1}}{\mathrm{4}\left({n}+\mathrm{1}\right)}\right)=\frac{\mathrm{5}}{\mathrm{24}} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{8}}\right)+\left(\frac{\mathrm{1}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{12}}\right)+\left(\frac{\mathrm{1}}{\mathrm{12}}−\frac{\mathrm{1}}{\mathrm{16}}\right)+ \\ $$$$\:\:\centerdot\centerdot\centerdot+\left(\frac{\mathrm{1}}{\mathrm{4}\left({n}−\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{4}{n}}\right)+\left(\frac{\mathrm{1}}{\mathrm{4}{n}}−\frac{\mathrm{1}}{\mathrm{4}\left({n}+\mathrm{1}\right)}\right)=\frac{\mathrm{5}}{\mathrm{24}} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}\left({n}+\mathrm{1}\right)}=\frac{\mathrm{5}}{\mathrm{24}}\:\Rightarrow\frac{\mathrm{1}}{\mathrm{4}\left({n}+\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{24}}\:\Rightarrow{n}+\mathrm{1}=\mathrm{6},\:{n}=\mathrm{5} \\ $$

Commented by hardmath last updated on 14/Oct/24

$$\mathrm{dear}\:\mathrm{professor}, \\ $$$$ \\ $$I don't quite understand this way, is there a shortcut for this?

Commented by Ar Brandon last updated on 14/Oct/24

Hello! Which line don't you understand?

Commented by hardmath last updated on 14/Oct/24

$$\mathrm{dear}\:\mathrm{professor}, \\ $$$$ \\ $$I do not fully understand this way, is there any other alternative way?

Commented by Ar Brandon last updated on 14/Oct/24

$${S}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{4}{n}}\:\Rightarrow{S}_{{n}+\mathrm{1}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{4}\left({n}+\mathrm{1}\right)}={S}_{{n}} −\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}\left({n}+\mathrm{1}\right)} \\ $$$$\Rightarrow\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{4}{n}}−\frac{\mathrm{1}}{\mathrm{4}\left({n}+\mathrm{1}\right)}\right)={S}_{{n}} −{S}_{{n}+\mathrm{1}} ={S}_{{n}} −\left({S}_{{n}} −\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}\left({n}+\mathrm{1}\right)}\right) \\ $$$$\Rightarrow\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{4}{n}}−\frac{\mathrm{1}}{\mathrm{4}\left({n}+\mathrm{1}\right)}\right)=\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}\left({n}+\mathrm{1}\right)}=\frac{\mathrm{5}}{\mathrm{24}},\:{n}=\mathrm{5} \\ $$

Commented by Ar Brandon last updated on 14/Oct/24

$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{4}{n}}−\frac{\mathrm{1}}{\mathrm{4}\left({n}+\mathrm{1}\right)}\right)=\frac{\mathrm{5}}{\mathrm{24}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{4}}−\cancel{\frac{\mathrm{1}}{\mathrm{8}}} \\ $$$$\:\:\:+\cancel{\frac{\mathrm{1}}{\mathrm{8}}}−\cancel{\frac{\mathrm{1}}{\mathrm{12}}} \\ $$$$\:\:\:+\cancel{\frac{\mathrm{1}}{\mathrm{12}}}−\cancel{\frac{\mathrm{1}}{\mathrm{16}}} \\ $$$$\:\:\:+\vdots \\ $$$$\:\:\:+\cancel{\frac{\mathrm{1}}{\mathrm{4}\left({n}−\mathrm{1}\right)}}−\cancel{\frac{\mathrm{1}}{\mathrm{4}{n}}} \\ $$$$\:\:\:+\cancel{\frac{\mathrm{1}}{\mathrm{4}{n}}}−\frac{\mathrm{1}}{\mathrm{4}\left({n}+\mathrm{1}\right)} \\ $$$$−−−−−−−−−− \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}\left({n}+\mathrm{1}\right)}=\frac{\mathrm{5}}{\mathrm{24}}\:\Rightarrow{n}=\mathrm{5} \\ $$

Commented by hardmath last updated on 14/Oct/24

$$\boldsymbol{\mathrm{T}}\mathrm{hank}\boldsymbol{\mathrm{Y}}\mathrm{ou}\:\boldsymbol{\mathrm{D}}\mathrm{ear}\boldsymbol{\mathrm{P}}\mathrm{rofessor} \\ $$

Answered by BaliramKumar last updated on 15/Oct/24

$$\frac{\mathrm{9}}{\mathrm{4}}\left[\frac{\mathrm{1}}{\mathrm{1}\centerdot\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{2}\centerdot\mathrm{3}}\:+\:…..\:+\:\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)}\right]\:=\:\frac{\mathrm{15}}{\mathrm{8}}\:\:\:\:\: \\ $$$$\left[\frac{\mathrm{1}}{\mathrm{1}\centerdot\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{2}\centerdot\mathrm{3}}\:+\:…..\:+\:\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)}\right]\:=\:\frac{\mathrm{15}}{\mathrm{8}}×\frac{\mathrm{4}}{\mathrm{9}}\:\:\:\:\:\: \\ $$$$\frac{\mathrm{2}−\mathrm{1}}{\mathrm{1}\centerdot\mathrm{2}}\:+\:\frac{\mathrm{3}−\mathrm{2}}{\mathrm{2}\centerdot\mathrm{3}}\:+\:…..\:+\:\frac{\left({n}+\mathrm{1}\right)−{n}}{{n}\left({n}+\mathrm{1}\right)}\:=\:\frac{\mathrm{5}}{\mathrm{6}}\:\:\:\:\:\: \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}\right)\:+\:…..\:+\:\left(\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)\:=\:\frac{\mathrm{5}}{\mathrm{6}} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}}−\cancel{\frac{\mathrm{1}}{\mathrm{2}}}+\cancel{\frac{\mathrm{1}}{\mathrm{2}}}−\cancel{\frac{\mathrm{1}}{\mathrm{3}}}\:+\:….\:+\:\cancel{\frac{\mathrm{1}}{{n}}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\:=\:\frac{\mathrm{5}}{\mathrm{6}} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\:=\:\frac{\mathrm{5}}{\mathrm{6}} \\ $$$$\frac{{n}}{{n}+\mathrm{1}}\:=\:\frac{\mathrm{5}}{\mathrm{6}} \\ $$$${n}\:=\:\mathrm{5} \\ $$

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