2 years ago

Answer 3+3 for 100 points brainlyist to first answer

Mathematics

2 answers:

jok3333 [9.3K]2 years ago

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Furkat [3]2 years ago

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

Flura [38]

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:

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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,

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

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