<s>[INST] Prove by mathematical induction that $3^n>2n^3$ I'm having trouble with this question:
"Prove by mathematical induction that for all integers $n\ge 6$, $3^n>2n^3$".
I got to $P(k)=2k^3<3^k$ and $P(k+1)=2(k+1)^3<3^{k+1}=2k^3+6k^2+6k+2<3^k*3$,
but I dont know how I can get $P(k+1)$ from $P(K)$...
Thanks
 [/INST] Suppose $2k^3<3^k$. Then
\begin{align}
2(k+1)^3&=2(k^3+3k^2+3k+1)\\
&=2k^3+6k^2+6k+2\\
&<3^k+6k^2+6k+2\\
&< 3^k+k^3+k^2+k\\
&<3^k+4k^3\\
&<3^k+2\cdot3^k\\
&=3^{k+1}
\end{align}
Note that we use in the middle that $6\le k$.
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