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natta225 [31]
3 years ago
15

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

Mathematics
1 answer:
Flura [38]3 years ago
3 0

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.

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