Question

5. (6 marks) Use mathematical induction to prove that for each integer n ≥ 4, 5^n ≥ 2^2n+1 + 100.

Answer 1

Step-by-step explanation:

We will prove by mathematical induction that, for every natural $n\geq 4$,  

$5^n\geq 2^{2n+1}+100$

We will prove our base case, when n=4, to be true.

Base case:

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

Inductive hypothesis:  

Given a natural $n\geq 4$,  

$5^n\geq 2^{2n+1}+100$

Now, we will assume the induction hypothesis and then use this assumption, involving n, to prove the statement for n + 1.

Inductive step:

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

With this we have proved our statement to be true for n+1.  

In conlusion, for every natural $n\geq4$.

$5^n\geq 2^{2n+1}+100$