Question

Show that 4n + 3^n > 6n, for all integers, n > 2.

Answer 1

Answer:

Induction.

Step-by-step explanation:

I will prove it with induction. For n=3:

$4(3)+3^{3} = 12 + 27 = 39 > 18.$

Suppose that the inequality is true for a k>3, that is to say

$4k+3^{k}>6k$

Then,

$4(k+1)+3^{k+1} = 4k+4+3^{k}3 = 4k+4+3^{k}+3^{k}+3^{k}$

$4k+3^{k}+4+3^{k}+3^{k} > 6k +4+1+1$ because $4k+3^{k}>6k$ and $3^{k}>1$ because k>0.

Then,

$4k+3^{k}+4+3^{k}+3^{k} > 6k+6$

$4k+3^{k}+4+3^{k}+3^{k} > 6(k+1)$

$4(k+1)+3^{k+1}> (6k+1)$.

In conclusion, for all integers n> 2 the inequality is true.