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Question Number 88440 by M±th+et£s last updated on 10/Apr/20

prove Σ_(k=1) ^∞ ((4^k −3^k )/(12^k ))=(1/6)

$${prove}\:\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{4}^{{k}} −\mathrm{3}^{{k}} }{\mathrm{12}^{{k}} }=\frac{\mathrm{1}}{\mathrm{6}} \\ $$

Commented by Tony Lin last updated on 10/Apr/20

((1/3)/(1−(1/3)))−((1/4)/(1−(1/4)))=(1/6)

$$\frac{\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}}−\frac{\frac{\mathrm{1}}{\mathrm{4}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$

Commented by M±th+et£s last updated on 10/Apr/20

thanx for solutions

$${thanx}\:{for}\:{solutions}\: \\ $$

Answered by Kunal12588 last updated on 10/Apr/20

Σ_(k=1) ^∞ (1/3^k )−Σ_(k=1) ^∞ (1/4^k )=((1/3)/(1−(1/3)))−((1/4)/(1−(1/4)))=(1/2)−(1/3)=(1/6)

$$\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{3}^{{k}} }−\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{4}^{{k}} }=\frac{\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}}−\frac{\frac{\mathrm{1}}{\mathrm{4}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}}=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$

Answered by mahdi last updated on 10/Apr/20

Σ_(i=1) ^∞ ((1/3))^i =(1/3)+(1/9)+...=(1/2) Σ_(i=1) ^∞ ((1/4))^i =(1/4)+(1/(16))+...=(1/3) Σ_(k=1) ^∞ ((4^k −3^k )/(12^k ))=Σ_(k=1) ^∞ ((1/3))^k −Σ_(k=1) ^∞ ((1/4))^k =(1/2)−(1/3)=(1/6)

$$\underset{\mathrm{i}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{i}} =\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{9}}+...=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\underset{\mathrm{i}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{i}} =\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{16}}+...=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{4}^{\mathrm{k}} −\mathrm{3}^{\mathrm{k}} }{\mathrm{12}^{\mathrm{k}} }=\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{k}} −\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{k}} =\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$ \\ $$

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