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Prove that 3^n is greater than n^2 for all n greater or equal to 0. I know that this proof will be by induction
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\[3^{n}>n ^{2},\forall n \in \mathbb{N} \]
hmmm its a while since I did these but here goes: if n = 1 we have 3^1 > 1^2 3 > 1 which is evidently true assuming the inequality is true then if n = k then 3^k > k^2 if n = k+1 then 3^(k+1) > (k + 1)^2 so the inequality is true for n=k then its also true for n = k+1 we have shown thats its treu for n = 1 therefore it must be true for n =2 , n = 3 ,4,5 etc so it must be true for all n>0
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