Question Number 88440 by M±th+et£s last updated on 10/Apr/20
$${prove}\:\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{4}^{{k}} −\mathrm{3}^{{k}} }{\mathrm{12}^{{k}} }=\frac{\mathrm{1}}{\mathrm{6}} \\ $$
Commented by Tony Lin last updated on 10/Apr/20
$$\frac{\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}}−\frac{\frac{\mathrm{1}}{\mathrm{4}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$
Commented by M±th+et£s last updated on 10/Apr/20
$${thanx}\:{for}\:{solutions}\: \\ $$
Answered by Kunal12588 last updated on 10/Apr/20
$$\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{3}^{{k}} }−\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{4}^{{k}} }=\frac{\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}}−\frac{\frac{\mathrm{1}}{\mathrm{4}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}}=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$
Answered by mahdi last updated on 10/Apr/20
$$\underset{\mathrm{i}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{i}} =\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{9}}+…=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\underset{\mathrm{i}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{i}} =\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{16}}+…=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{4}^{\mathrm{k}} −\mathrm{3}^{\mathrm{k}} }{\mathrm{12}^{\mathrm{k}} }=\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{k}} −\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{k}} =\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$ \\ $$