We want to prove by induction that
3n>2n2 Base case: n=1;
31>2(1)2⟹3>2 Thus S1 is true.
Inductive Step:
We assume it is true for n=k, so that
3k>2k2
Inductive Step:
We will show that since it is true for n= k, then it is true for n=k+1.
3k+1=3k⋅3>2k2⋅3>6k2>2k2+4k2>2[(k+1)2−2k−1]+4k2>2(k+1)2−4k−2+4k23k+1>2(k+1)2+4k2−4k−2
Since
3k+1>2(k+1)2+4k2−4k−2 then
3k+1>2(k+1)2 which concludes the proof.
Comments