Prove by Mathematical Induction that $3^n > n^2 + 4$ for $n\geq 2$.

Question

I end up getting this as the last step: $3^{k+1} > (k+1)^2 + 4$

$3^k * 3 > (k^2 + 4) + 2k + 1$

and here is where I get stuck.

Answer 1

If your question is to prove that $3^{n} >n^{2}+4$ note that this is true for $n=2$. If it holds for some $n$ then $3^{n+1} =3(3^{n}) >3(n^{2}+4)$. We have to show that $3n^{2}+12 >(n+1)^{2}+4$ or $2n^{2}-2n+7>0$. write LHS as $2(n-\frac 1 2)^{2}+(7-\frac 1 2)$ which is clearly positive.

Answer 2

By the binomial theorem, if $n\ge 3$ then $$ 3^n=(2+1)^n = 2^n + n2^{n-1} + n(n-1)2^{n-3} + \cdots \ge 2^n + 2^{n - 3} (n^2 + 3 n) \ge 2^n + 2^{n - 3} n^2 \ge 4 + n^2 $$

Answer 3

You have done the half of the proof you need just to show that : $$3^{n+1}>(n+1)^2+4$$ By the induction hypothesis you have : $$3^n>n^2+4$$ Thus : \begin{align} 3^n\times 3&>3(n^2+4) \\\ 3^{n+1}&>3n^2+12 \end{align} Now we need to show that : $$3n^2+12>(n+1)^2+4$$ The best way to do this is by the difference :

\begin{align} 3n^2+12-\bigg((n+1)^2+4\bigg)&=3n^2+12-n^2-2n-5 \\\ &=2n^2-2n+7\\\ &=n(2n-2)+7 \end{align} Since $n \ge 2$ thus : \begin{align} 2n &\ge 4 \\\ 2n-2 &\ge 2 \\\ n(2n-2)& \ge 4 \end{align} Finally $n(2n-2)+7$ is positive : Therefore : $$3n^2+12>(n+1)^2+4$$ And : $$3^{n+1}>(n+1)^2+4$$ Hence by induction we have shown that : $$3^n>n^2+4$$ is a true property for all $n\ge 2$.

I hope that my answer is clear, Good luck !