Can Induction Prove the Inequality 2n + 1 < 2^n for n ≥ 3?

[killerfish]

Homework Statement

Prove: 2n + 1 < 2n , with n >= 3

Homework Equations

The Attempt at a Solution

2 (3) + 1 = 7 and 2³ = 8.
So 2 (3) + 1 < 2³.
Thus the inequality holds with n = 3:
Suppose the inequality holds with n = k
Then 2k+ 1 < 2^k:
So 2k + 1 + 2 < 2^k + 2
2k + 3 < 2^k + 2^k
2k + 3 < 2(2^k)
2 (k + 1) + 1 < 2^(k+1):
So, the inequality holds with n = k + 1:
Hi guys,

some of the transition on the RHS, I am blurred. Like the above there are 2 parts i don't understand,

So 2k + 1 + 2 < 2^k + 2
2k + 3 < 2^k + 2^k

on RHS, how to get from 2^k+2 to 2^k+2^k. Arent when we do a change on LHS(e.g +2), is should be equal to RHS(e.g+2)? sry my understanding for induction is weak, can someone help elaborate the solution...

Thanks very much.

 

[kuruman]

It's really simple.

You already have that the LHS is less than the RHS

2k + 3 < 2^k + 2

If you replace the 2 on the RHS with 2^k, a number greater than 2 (k > 1 by assumption), the inequality still holds, therefore

2k + 3 < 2^k + 2^k.

 

[HallsofIvy]

2< 2^k for any k> 1 and so, adding 2^k to both sides, 2^k+ 2< 2^k+ 2^k so 2^k+ 2< 2(2^k)= 2^k+1.