Question

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

Answer 1

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

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`.