After proving the above inequality is true for n=1, I assume P(k): 3^k>2k^2-1 is true, and P(k+1):3(3^k)>3(2k^2-1). Then I'd need to prove that 3(2k^2-1) is greater than 2(k+1)^2-1. If I work backwards and do expand along these lines, I end up with a quadratic equation k^2-k-1>0 where the ranges k>1.618 and k<-0.618 satisfy the aforementioned inequality. However, since k is defined to be greater than 1, this does not compute for k>1.618.
Prove that 3^n>2n^2-1 for all positive integers n via mathematical induction.
I'm guessing since I've proven that P(n) is true for n=1 earlier, and since by definition n is an integer (hence the next integer value for n would be 2), hence k is true for 2 since 2>1.618. Is this necessarily true or am I missing something here?