Can you use mathematical induction to prove that $2^{n} > 4 n$ $2^n > 4n$ for all $n in ZZ^+ n>=5$ ?

Question

15415 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-03T06:38:57

Proof: (by induction)

Base case: For $n=5$ , we have $2^5 = 32 > 20 = 4(5)$ .

Inductive hypothesis: Suppose that $2^k > 4k$ for some integer $k>=5$ .

Induction step: We wish to show that $2^(k+1) > 4(k+1)$ . Indeed,

$2^(k+1) = 2*2^k$

$=2^k+2^k$

$>4k+4k" "$ (by the inductive hypothesis)

$>4k+4" "$ (as $k>=5$ )

$=4(k+1)$

We have supposed true for $k$ and shown true for $k+1$ . Thus, by induction, $2^n>4n$ for all integers $n>=5$ .
∎

Can you use mathematical induction to prove that 2^n > 4n2n>4n2^n > 4n for all n in ZZ^+ n>=5n in ZZ^+ n>=5?