Use Mathematical Induction to prove that for all integers $n\geq 5, 1+4n<2^n$

Question

So far I have gotten this far:

Proof: Let $S(n)$ be the statement above $S(n)=n\geq 5,1+4n<2^n$

      Now let us prove base case of n=5, S(5)=5≥5,1+4(5)<2^5=21<32.
      Let us assume that for S(k)is the statement 1+4k <2^k is true for k≥5.
      S(k+1) is the statement  1+4(k+1) <2^(k+1)=4k+5<2^(k+1)

The problem is I don't know where to go after this. Please help me finish my inductive step.

Answer 1

You have $1+4k<2^k$, and you want to show that $1+4(k+1)<2^{k+1}$.

If you can show you have added more to the right side than to the left side, you're golden.

What you have added to the left side is $1+4(k+1)-(1+4k)=4$.

What you have added to the right side is $2^{k+1}-2^k=2^k$.

Can you take it from here?

Answer 2

Just note that $1+4(k+1)=1+4k+4 < 2^k + 4 \le 2^k + 2^k = 2^{k+1}$ because $4 \le 2^k$ for $k\ge 2$.