Question

Use mathematical induction to prove that for each integer n ≥ 4, 5^n ≥ 2^2n+1 + 100. (it is 5 to power of n and 2 to the power of 2n+1)

Answer 1

Step-by-step explanation:

The statement to be proved using mathematical induction is:

1. "For every $n\geq 4$, $5^n\geq 2^{2n+1}+100$

We will begin the proof showing that the base case is satisfied (n=4).

$5^4=625\geq 612=2^{2*4+1}+100$.

Then, 1 is true for n=4.

Now we will assume that the statement holds for some arbitrary natural number $n\geq 4$ and prove that then, the statement holds for n+1. Observe that

$2^{2(n+1)+1}+100=2^{2n+1+2}+100=4*2^{2n+1}+100\leq 4(2^{2n+1}+100)\leq 4*5^n$

With this the inductive step has been proven and then, our statement is true,

For every $n\geq 4$, $5^n\geq 2^{2n+1}+100$