Answer to Question #203289 in Algebra for Raghav

Question

Prove that 2ⁿ>4n for n≥5.

Answer 1

Prove for $n=5$ :

$2^5=32>20=4\cdot5$

We prove by induction, let be true for $n-1$

$2^n=2\cdot2^{n-1}\stackrel{\text{True for n-1}}{>}2\cdot4(n-1)=4n+(4n-8)$

Because $4n-8>0$ for $n\geq5$ :

$4n+(4n-8)>4n$

So, we prove that $2^n>4n$ for $n\geq5$