Prove using induction that $(2^n)+1<3^n$ for all natural numbers, $n>=2$

Question

I needed help proving this statement, this is what I have tried so far

Base case:

$n = 2$

$5 < 9$ $->$ $True$

Inductive step:

Assume that for some $k>=2$ , $(2^k)+1<3^k$ show that $P(k+1)$ holds

-> 2^(k+1) + 1
-> 2*(2^k) + 1
-> (1+1)*(2^k) + 1
-> 2^k + 2^k + 1
 <  2^k + 3^k

This is where I get stuck I am not sure where to go from there or how to manipulate that to get 3^(k+1)

Any help would be appreciated thank you

Answer 1

Firs note that $2<3$ implies $3/2 > 1$.

Note how the $2^k$ sequence grows: $2^{k+1}= 2^k\times 2$. That is, multiply previous term by $2$.

Now look at $3^{k+1} = 3^k \times 2 \times (3/2)$. This multiplies by $2$ and further multiplies by $3/2$.

At $k=2$ the two sequences differ by $9-4=5$ So, the difference of 4 will be maintained. (If you use calculus you can say something much stronger). That is $2^k+ 4 < 3^k$ for $k\ge 2$.

Answer 2

If you are using induction, assume $2^k+1<3^k$, then multiply by 2, $2(2^k+1)<2*3^k$, or $2^{k+1}+2<2*3^{k}$, and since $2*3^k-1<3*3^k$ then $2^{k+1}+1<3^{k+1}$, and the induction is complete