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Archangel Macsika

Let n be a natural number. Use mathematical induction to prove that 4 n−1 > n2 for all n ≥ 3.

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Let the statement P(n) = 4^{n−1} > n^2

Now, 4^{3-1} = 4^2 = 16 > 3^2 , So P(n) is true for n = 3 .

Let P(n) is true for n = k \implies 4^{k-1} > k^2 .

Now, prove P(n) is true for n = k+1.

4^{(k+1)-1} = 4 \times 4^{k-1} > 4 \times k^2 > k^2 + 2k^2+k^2 > k^2 + 2k +1 because k \geq 3 .

Hence, 4^{(k+1)-1} > (k+1)^2

Hence, using Principal of Mathematical Induction, 4^{n-1}>n^2 for all n\geq 3 .

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