Question #241156

By using principle of mathematical induction prove 2^n > n^2

if n is an integer greater than 4.


1
Expert's answer
2021-09-24T00:59:12-0400

Let P(n)P(n) be the proposition that 2n>n2,n>4.2^n > n^2, n>4.

BASIS STEP: P(5)P(5) is true, because 25=32>25=52.2^5=32 > 25=5^2.


INDUCTIVE STEP: For the inductive hypothesis we assume that P(k)P(k) holds for an arbitrary integer k>4.k>4. That is, we assume that


2k>k2,k>4.2^k > k^2, k>4.

Under this assumption, it must be shown that P(k+1)P(k + 1) is true, namely, that


2k+1>(k+1)22^{k+1} > (k+1)^2

We have that


2k+1=2(2k)>2k2=k2+k2=k2+k(k)2^{k+1}=2(2^{k})>2k^2=k^2+k^2=k^2+k(k)

>k2+k(2+1)>k2+2k+1=(k+1)2,k>4>k^2+k(2+1)>k^2+2k+1=(k+1)^2, k>4

This shows that P(k+1)P(k + 1) is true under the assumption that P(k)P(k) is true. This completes the inductive step

We have completed the basis step and the inductive step, so by mathematical induction we know that P(n)P(n) is true for all integers n>4.n>4.

That is, we have proven that 2n>n22^n > n^2 for all integers n>4.n>4.



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