Step 1) Prove for n=8
LHS:2^8=256 RHS: 3(8)^2=192, therefore 2^n>3n^2 is true for n=8
Step 2) Assume true for n=k
ie 2^k>3k^2
Step 3) Prove for n=k+1
from step 2, 2^k>3k^2
therefore, 2(2^k)>2(3k^2)
therefore, 2^(k+1)>6k^2
therefore, 2^(k+1)>3(k+1)^2+3k^2-6k-3>3(k+1)^2 [as (3k^2-6k-3) is...