Question

prove that 2^n > n for all n>=2

Answer 1

this means that \[2^{n+1} = 2^{n}\times n > .... n+1\]

Answer 2

to me, I will use induction to prove
basic case n =1 2^1 >1 true
hypothesis step: n =k , 2^k >k true
need to prove n = k+1 true. It means \(2^{k+1}> k+1\)
from hypothesis step time both sides by 2
\(2^k *2 > 2k \) since k>2 , so 2k> k+1, therefore \(2^{k+1}>2k>k+1\) or \(2^{k+1}>k+1\)

Answer 3

can you explain to me better how the 2k > k+1 is true

Answer 4

take k =2 , 2*2 > 2+1, right?

Answer 5

one more, k =3 so 2*3 > 3+1, right?

Answer 6

that's it

Answer 7

Actually 2^n > n holds true for all values of n