can anybody help me for this question?..i alrdy used the sum to infinity formula..but may be my first term and common ratio r wrong...so i cannot get the correct answer, the answer given is 1
Sent from my Lenovo A5000 using Tapatalk
can anybody help me for this question?..i alrdy used the sum to infinity formula..but may be my first term and common ratio r wrong...so i cannot get the correct answer, the answer given is 1
Sent from my Lenovo A5000 using Tapatalk
Let $y = \sqrt[3]{4\sqrt[3]{4\sqrt[3]{4 \, ... }}} \implies y=\sqrt[3]{4y} \implies y^3 = 4y \implies y=2$ since $y>0$
Let $x= \sqrt{2\sqrt{2\sqrt{2 \, ... }}} \implies x=\sqrt{2x} \implies x^2 = 2x \implies x=2$ since $x>0$
$\dfrac{y}{x} = \dfrac{2}{2} = 1$