Prove by mathematical induction that $3^n>2n^3$

Question

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

Answer 1

$P(k)$ is the statement $ 2n^3<3^n $. Do not write "$P(k)=.....$"; $P(k)$ is not a mathematical value.

If we assume that then we have $2n^3 < 3^n$

So $2(n+1)^3 = 2n^3 + 6n^2 + 6n + 1$ And we have $2n^3 < 3^n$ so

$2(n+1)^3 =2n^3 + 6n^2 + 6n + 1< 3^n + 6n^2 + 6n + 1$

And $n\ge 6$ so $6n^2 \le n*n^2 =n^3$ and $6n+1 < 6n+n < 6n*n=6n^2 < n*n^3 < n^3$.

So

$2(n+1)^3 =2n^3 + 6n^2 + 6n + 1< 3^n + (6n^2) + (6n + 1)$

$< 3^n + n^3 + n^3 = 3^n + 2n^3 \le 3^n + 3^n < 3^n + 3^n + 3^n$

$< 3*3^n = 3^{n+1}$.

So $2(n+1)^3 < 3^{n+1}$ and so the statement $P(k+1)$ is true.

Answer 2

Note that $f(k) = \frac{k}{k+1}$ is a positive, strictly increasing function for positive $k$ (since $\frac{k}{k+1} = 1 - \frac{1}{k+1}$ and $\frac{1}{k+1}$ is strictly decreasing). Thus, for $k \ge 6$, you have

$$\begin{equation}\begin{aligned} f^3(k) & = \left(\frac{k}{k+1}\right)^3 \\ & \ge \left(\frac{6}{7}\right)^3 \\ & = \frac{216}{343} \\ & \gt \frac{1}{3} \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

Using this, you get with the induction part that

$$\begin{equation}\begin{aligned} 3^{k+1} & = 3(3^{k}) \\ & > 3(2k^3) \\ & = 2(3)\left(\frac{k}{k+1}\right)^3(k+1)^3 \\ & \gt 2(3)\left(\frac{1}{3}\right)(k+1)^3 \\ & = 2(k + 1)^3 \end{aligned}\end{equation}\tag{2}\label{eq2A}$$

Thus, this shows that if $P(k)$ is true, then so is $P(k+1)$.

Answer 3

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 $6\le k$.

Answer 4

$$P(k+1)=2(k+1)^3<\left(\dfrac{k+1}k\right)^33^k$$

So, it is sufficient to establish $\left(\dfrac{k+1}k\right)^3<3$

which is true if $\dfrac1k<\sqrt[3]3-1\iff k>\dfrac1{\sqrt[3]3-1}$

Now $3>\dfrac1{\sqrt[3]3-1}\iff\dfrac13+1<\sqrt[3]3\iff\dfrac{64}{27}<3\iff64<81$ which is true

So, $\left(\dfrac{k+1}k\right)^3<3$ if $k\ge3$

Now establish $P(6)$

Answer 5

We have to prove that for $n \geq 6$ it is $$ 3^{h + 1} > 2\left( {h + 1} \right)^3 $$ This is equivalent to $$ 3^h + 3^h + 3^h > 2h^3 + 6h^2 + 6h + 2 $$ We have that

1. $3^h>2h^3$ by the inductive hypothesis.
2. $3^h>2h^3>h^3\geq 6h^2$ by the inductive hypothesis and since $h \geq 6$.
3. $3^h>2h^3>h^3>6h+2$ since $$ h^3 - 6h = h\left( {h^2 - 6} \right) > 2 $$ as long as $h^2-6>0$. Namely, in this case it is $h^2-6h \geq 1$ and $h(h^2-6h) \geq 6$