Question

prove 1+ 2^n < 3^n for n>=2

Answer 1

induction step

Answer 2

Is it true for \(n=2\) (the base case)?
\[1+2^2=5<9=3^2\]
Yes.

Assume it's true for \(n=k\):
\[1+2^k<3^k\]
Use this to show it's true for \(n=k+1\). What you want to show:
\[(*)~~~~1+2^{k+1}<3^{k+1}\]
Rewriting \((*)\), you could have
\[1+2(2^k)=\color{red}{1+2^k}+2^k<\color{red}{3^k}+2^k\]
The red part is true by the induction hypothesis. We need to establish that \(3^k+2^k\le3^{k+1}\).
\[3^{k+1}=3(3^k)=3^k+3^k+3^k\]
If you can accept that \(2^k<3^k\) (you might have to prove) for all \(k\ge1\), you're done, since
\[\color{red}{3^{k+1}}=3^k+3^k+3^k\color{red}>3^k+2^k+2^k=\color{red}{3^k+2^{k+1}>3^k+2^k>1+2^{k+1}}\]

Answer 3

Thanks.

Answer 4

You're welcome