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

\frac{1 + 2 + 2^{2} + 2^{3} + \dots + 2^{n - 1}}{1 + 2 + 2^{2} + 2^{3} + \dots + 2^{n}} and

\frac{1 + 3 + 3^{2} + 3^{3} + \dots + 3^{n - 1}}{1 + 3 + 3^{2} + 3^{3} + \dots + 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} + \dots + 2^{n - 1}}{1 + 2 + 2^{2} + 2^{3} + \dots + 2^{n}} = \frac{2^{n - 1} - 1}{2^{n} - 1} \dots (i)

\frac{1 + 3 + 3^{2} + 3^{3} + \dots + 3^{n - 1}}{1 + 3 + 3^{2} + 3^{3} + \dots + 3^{n}} = \frac{3^{n - 1} - 1}{3^{n} - 1} \dots . (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)