Prove that P:2^n>2n+1 for integers n>=3

[Lis]

I have to do this by induction, I know I have to prove it for P(3)
then suppose it is true por P(k)
then 2^k>2k+1 is true this'd be the induction hypothesis
and I have to prove that P(k+1) is true

Then:
2^(k+1)>2(k+1)+1
(2)2^k>2k+2+1
(2)2^k>(2k+1)+2

from then on I don't know what to do to prove it's true can someone help me please?

 

[Deleted member 4993]

I have to do this by induction, I know I have to prove it for P(3)
then suppose it is true por P(k)
then 2^k>2k+1 is true this'd be the induction hypothesis
and I have to prove that P(k+1) is true

Then:
2^(k+1)>2(k+1)+1
(2)2^k>2k+2+1
(2)2^k>(2k+1)+2

from then on I don't know what to do to prove it's true can someone help me please?

Assume

2^(k) > 2k +1

Then

2^(k+1)

2*2^k → > 2*(2k+1) = 4k +2 = 2k + (2k+2) > 2k + 3 = 2*(k+1) + 1

Hence

2^(k+1) > 2(k+1) + 1