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?


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

$Which of the two numbers$ $\frac{1 + 2 + 2^{2} + 2^{3} + . . . + 2^{n - 1}}{1 + 2 + 2^{2} + 2^{3} + . . . + 2^{n}} and$ $\frac{1 + 3 + 3^{2} + 3^{3} + . . . + 3^{n - 1}}{1 + 3 + 3^{2} + 3^{3} + . . . + 3^{n}} is greater ?$
Commented by danger last updated on 05/Apr/20

$1 s t o n e$
Commented by danger last updated on 05/Apr/20

$n o t c l e a r$
Commented by $@ty@m123 last updated on 05/Apr/20

${I t}^{'} s b e t t e r i f y o u t y p e t h e q u e s t i o n .$
Commented by danger last updated on 05/Apr/20

$y o u r q u e s t i o n i s n o t c o m p l e t e$
Commented by danger last updated on 05/Apr/20

$o k d o n e$
	Answered by $@ty@m123 last updated on 05/Apr/20

	$\frac{1 + 2 + 2^{2} + 2^{3} + . . . + 2^{n - 1}}{1 + 2 + 2^{2} + 2^{3} + . . . + 2^{n}} = \frac{2^{n - 1} - 1}{2^{n} - 1} . . . (i)$ $\frac{1 + 3 + 3^{2} + 3^{3} + . . . + 3^{n - 1}}{1 + 3 + 3^{2} + 3^{3} + . . . + 3^{n}} = \frac{3^{n - 1} - 1}{3^{n} - 1} . . . . (i i)$ $P u t n = 2, t h e n$ $(i) \equiv \frac{1}{3} & (i i) \equiv \frac{2}{8} = \frac{1}{4}$ $H e n c e (i) > (i i)$
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