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Prove that 2n+1 <= 2^n for n >= 3 By induction: Proved P(3) is true Inductive step: P(n+1): 2n+3 <= 2^(n+1) 0 <= 2(2^n) - 2n + 3 I'm stuck trying to prove this part, help?
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Its actually 0 <= 2(2^n) - 2n - 3
In the inductive step, you can use the fact that \(2n+1 \le 2^n\). \(2n+3=(2n+1)+2\le 2^n+2\). if by any chance \(2^n+2<2^{n+1}\) (for n>=3) then you are done.
Assume for some \(n\) we have \(P(n)\colon 2n+1\le 2^n\). Now observe:$$2n+1\le 2^n\\2n+1+2\le 2^n+2\\2(n+1)+1\le 2^n+2\le 2^{n+1}$$
\(2^n+2<2^{n+1}\) trivially... think of binary :-p
I'm confused as to how it absolutely proves that 2^n+1 is actually greater for all n
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