Let the statement "P(n) = 4^{n\u22121} > 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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