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$4^{n-1}> n^2$ , for all integers $n\geq 3$. I have started it in this way: Let $P(n): 4^{n-1}>n^2$, for all integers $n\geq 3$. To show $P(3)$ is true: $4^{3-1}>3^2$ $16>9$, hence $P(3)$ is true. Suppose that $P(k)$ is true for all $n\geq 3$, that is $4^{k-1}>k^2$ We must show that $P(k+1)$ is also true $4^{(k+1)-1}>(k+1)^2$ Left hand side: $4^k=4^k × 4/4 =4^{k-1} ×4 >k^2 ×4$ But I'm not able to make it equal to right hand side, to prove the Induction.

mathma
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Areesha
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    "Suppose that $p(k)$ is true for some particular k that $k\geq 3$" You wrote "for all $n\geq 3$" here which is incorrect. Now, as for the induction step, note that $4\cdot 4^{k-1} = 4^{k-1}+4^{k-1}+4^{k-1}+4^{k-1}$, each of these you should be able to recognize. Now, recall that $k^2=k\cdot k \geq 2\cdot k\geq 1$ and that $(k+1)^2=k^2+2k+1$ – JMoravitz May 19 '21 at 19:28
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    The end result is that there are many different ways you can prove the induction step, so pick the one that makes the most sense to you. As for structuring your proof, I recommend reading How to write a clear induction proof – JMoravitz May 19 '21 at 19:30

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You are almost done. Assume that $P(k)$ is true. Then, $$4^{k-1}>k^2$$

Multiply both sides by $4$. $$4^k>4k^2=(2k)^2>(k+1)^2$$

h-squared
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