Use induction to prove that $2^n \gt n^3$ for every integer $n \ge 10$.

Question

Use induction to prove that $2^n \gt n^3$ for every integer $n \ge 10$.

My method:

If $n = 10$, $2^n \gt n^3$ where $2^{10} \gt 10^3$ which is equivalent to $1024 \gt 1000$, which holds for $n = 10$. $2^k \gt k^3$.

$2^{k + 1} \gt (k + 1)^3$

$2^{k + 1} \gt (k + 1)(k + 1)(k + 1)$

$2^{k + 1} \gt (k^2 + 2k + 1)(k + 1)$

$2^{k + 1} \gt (k^3 + 3k^2 + 3k + 1)$

$(k + 1) \cdot 2^k \gt k^3 + 3k^2 + 3k + 1$

Since $2^{10} \gt 10^3$, the inequality holds for $n = 10$. Assume $2^k \gt k^3$ for $k \ge 10$. We show that $2^{k + 1} \gt (k + 1)^3$. Hence, $2^{k + 1} = 2 \cdot 2^k \gt 2k^3$

$= k^3 + k^3 \ge k^3 + 10k^2$

$= k^3 + 10k^2 = k^3 + 4k^2 + 6k^2 \ge k^3 + 4k^2 + 6 \cdot 10 = k^3 + 4k^2 +60$

$k^3 + 4k^2 + 60 \gt k^3 + 3k^2 + 3k + 1$

Therefore, $2^{k + 1} \gt (k + 1)^3$. Hence, $2^n \gt n^3$ for every integer $n \ge 10$.

I was trying to fix this but I am not sure how to go about doing so.

Answer 1

We do the induction step, perhaps somewhat along the lines that you had considered.

Let $k\ge 10$, and suppose that $2^k\gt k^3$. We want to show that $2^{k+1}\gt (k+1)^3$.

We have $$2^{k+1}=2\cdot 2^k\gt k^3+k^3.$$ It will now be enough to show that $k^3+k^3\gt (k+1)^3$, or equivalently that $(k+1)^3\lt k^3+k^3$. We have $$(k+1)^3=k^3+3k^2+3k+1\le k^3+3k^2+3k^2+k^2=k^3+7k^2.$$ But $7k^2\lt k^3$, and therefore $(k+1)^3\lt k^3+k^3$.

Answer 2

Hint: if $n\geq 10$, $$ \frac{(n+1)^3}{n^3}=\frac{n^3+3n^2+3n+1}{n^3}=1+\frac 3n+\frac 3{n^2}+\frac 1{n^3}\leq 1+\frac 3{10}+\frac 3{100}+\frac 1{1000}<2. $$