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Question Number 87563 by Serlea last updated on 05/Apr/20

Which of the two numbers ((1+2+2^2 +2^3 +...+2^(n−1) )/(1+2+2^2 +2^3 +...+2^n )) and ((1+3+3^2 +3^3 +...+3^(n−1) )/(1+3+3^2 +3^3 +...+3^n )) is greater?

$$\mathrm{Which}\:\mathrm{of}\:\mathrm{the}\:\mathrm{two}\:\mathrm{numbers} \\ $$$$\frac{\mathrm{1}+\mathrm{2}+\mathrm{2}^{\mathrm{2}} +\mathrm{2}^{\mathrm{3}} +...+\mathrm{2}^{\mathrm{n}−\mathrm{1}} }{\mathrm{1}+\mathrm{2}+\mathrm{2}^{\mathrm{2}} +\mathrm{2}^{\mathrm{3}} +...+\mathrm{2}^{\mathrm{n}} }\:\mathrm{and}\: \\ $$$$\frac{\mathrm{1}+\mathrm{3}+\mathrm{3}^{\mathrm{2}} +\mathrm{3}^{\mathrm{3}} +...+\mathrm{3}^{\mathrm{n}−\mathrm{1}} }{\mathrm{1}+\mathrm{3}+\mathrm{3}^{\mathrm{2}} +\mathrm{3}^{\mathrm{3}} +...+\mathrm{3}^{\mathrm{n}} }\:\mathrm{is}\:\mathrm{greater}? \\ $$

Commented by danger last updated on 05/Apr/20

1st one

$$\mathrm{1}{st}\:{one} \\ $$

Commented by danger last updated on 05/Apr/20

not clear

$${not}\:{clear} \\ $$

Commented by $@ty@m123 last updated on 05/Apr/20

It′s better if you type the question.

$${It}'{s}\:{better}\:{if}\:{you}\:{type}\:{the}\:{question}. \\ $$

Commented by danger last updated on 05/Apr/20

your question is not complete

$${your}\:{question}\:{is}\:{not}\:{complete} \\ $$

Commented by danger last updated on 05/Apr/20

ok done

$${ok}\:{done} \\ $$

Answered by $@ty@m123 last updated on 05/Apr/20

((1+2+2^2 +2^3 +...+2^(n−1) )/(1+2+2^2 +2^3 +...+2^n )) =((2^(n−1) −1)/(2^n −1)) ...(i) ((1+3+3^2 +3^3 +...+3^(n−1) )/(1+3+3^2 +3^3 +...+3^n )) =((3^(n−1) −1)/(3^n −1)) ....(ii) Put n=2, then (i)≡(1/3) & (ii)≡ (2/8)=(1/4) Hence(i)>(ii)

$$\frac{\mathrm{1}+\mathrm{2}+\mathrm{2}^{\mathrm{2}} +\mathrm{2}^{\mathrm{3}} +...+\mathrm{2}^{\mathrm{n}−\mathrm{1}} }{\mathrm{1}+\mathrm{2}+\mathrm{2}^{\mathrm{2}} +\mathrm{2}^{\mathrm{3}} +...+\mathrm{2}^{\mathrm{n}} }\:=\frac{\mathrm{2}^{{n}−\mathrm{1}} −\mathrm{1}}{\mathrm{2}^{{n}} −\mathrm{1}}\:...\left({i}\right) \\ $$$$\frac{\mathrm{1}+\mathrm{3}+\mathrm{3}^{\mathrm{2}} +\mathrm{3}^{\mathrm{3}} +...+\mathrm{3}^{\mathrm{n}−\mathrm{1}} }{\mathrm{1}+\mathrm{3}+\mathrm{3}^{\mathrm{2}} +\mathrm{3}^{\mathrm{3}} +...+\mathrm{3}^{\mathrm{n}} }\:=\frac{\mathrm{3}^{{n}−\mathrm{1}} −\mathrm{1}}{\mathrm{3}^{{n}} −\mathrm{1}}\:....\left({ii}\right) \\ $$$${Put}\:{n}=\mathrm{2},\:{then} \\ $$$$\left({i}\right)\equiv\frac{\mathrm{1}}{\mathrm{3}}\:\:\&\:\left({ii}\right)\equiv\:\frac{\mathrm{2}}{\mathrm{8}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${Hence}\left({i}\right)>\left({ii}\right) \\ $$

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