Question 495865
The base cases (n=1, 2) hold. For some n>2, if


*[tex \LARGE 3^n > 1 + 2^n], then


*[tex \LARGE 3^{n+1} = 3(3^n)]


from our original inequality, we establish


*[tex \LARGE 3^{n+1} = 3(3^n) > 3(1+2^n)]


*[tex \LARGE  = 3 + 3(2^n) = (1 + 2^{n+1}) + (2 + 2^n)]


Hence, *[tex \LARGE 3^{n+1} > (1+2^{n+1})] and the induction is complete.