Question

Use mathematical induction to prove that for each integer n > 4,5" > 2^2n+1 + 100.

Answer 1

Answer:

The inequality that you have is $5^{n}>2^{2n+1}+100,\,n>4$. You can use mathematical induction as follows:

Step-by-step explanation:

For $n=5$ we have:

$5^{5}=3125$

$2^{(2(5)+1)}+100=2148$

Hence, we have that $5^{5}>2^{(2(5)+1)}+100.$

Now suppose that the inequality holds for $n=k$ and let's proof that the same holds for $n=k+1$. In fact,

$5^{k+1}=5^{k}\cdot 5>(2^{2k+1}+100)\cdot 5.$

Where the last inequality holds by the induction hypothesis.Then,

$5^{k+1}>(2^{2k+1}+100)\cdot (4+1)$

$5^{k+1}>2^{2k+1}\cdot 4+100\cdot 4+2^{2k+1}+100$

$5^{k+1}>2^{2k+3}+100\cdot 4$

$5^{k+1}>2^{2(k+1)+1}+100$

Then, the inequality is True whenever $n>4$.