Proving by induction: $2^n > n^3 $ for any natural number $n > 9$ [duplicate]

Solution 1:

For another way just using $n>9$, note that when $n=10$, $2^n = 1024 > 1000 = n^3$. Now suppose that $2^n>n^3$ for $n>9$. Then,

$\begin{align*} 2^{n+1} &= 2\cdot2^n \\ &>2n^3 \\ &= n^3 +n^3 \\ &> n^3 + 9n^2 \\ &= n^3 + 3n^2 + 6n^2 \\ &>n^3 + 3n^2 +54n \\ &=n^3+3n^2+3n +51n\\ &>n^3+3n^2+3n+1 \\ &= (n+1)^3. \end{align*}$

Solution 2:

For your "subproof":

Try proof by induction (another induction!) for $k \geq 7$

$$k^3 > 3k^2 + 3k + 1$$

And you may find it useful to note that $k\leq k^2, 1\leq k^2$

$$3k^2 + 3k + 1 \leq 3(k^2) + 3(k^2) + 1(k^2) = 7k^2 \leq k^3 \quad\text{when}??$$

Solution 3:

Here's another way. Suppose $k>9$. Then: $$ \begin{align*} k^3&=(k)k^2\\ &>9k^2\\ &=3k^2+3k^2+3k^2\\ &=3k^2+3(k)k+3(k)^2\\ &>3k^2+3(9)k+3(9)^2\\ &=3k^2+27k+243\\ &>3k^2+3k+1\\ \end{align*}$$

Solution 4:

If $P(n): 2^n>n^3$

$n=10, P(10): 2^{10}=1024$ and $10^3=1000$

Let $P(n)$ is true for $n=m\implies 2^m>m^3$

Now, $P(m+1): 2^{m+1}=2\cdot 2^m>2m^3$ which we need to be $>(m+1)^3$

$\implies 2>\left(1+\frac1m\right)^3 $

If $m=2, \left(1+\frac1m\right)^3=\frac{27}8>2$

If $m=3, \left(1+\frac1m\right)^3=\frac{64}{27}>2$

If $m=4, \left(1+\frac1m\right)^3=\frac{125}{64}<2$

$\implies 2>\left(1+\frac1m\right)^3$ for $m=4$

and $\left(1+\frac1m\right)^3>\left(1+\frac1{m+1}\right)^3\implies 2>\left(1+\frac1m\right)^3$ for $m\ge 4$