Menu Close

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-




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

Leave a Reply

Your email address will not be published. Required fields are marked *