Proof for Induction Question

Question

Hey having trouble finishing this question.

Prove by induction that $n^3 \le 2^n$ for all natural numbers $n\ge 10$.

This is what I have so far:

Base step: For $n = 10$

$1000 \le 1024$

Assumption Step: For $n = k$

Assume $k^3 \le 2^k$

Induction step: For $n = (k+1)$

$(k+1)^3 \le 10^{k+1}$

$k^3 +3k^2 + 3k +1 \le 10^k*10$

Not really sure where to go from here

Hint : if $n^3 \leq 2^n$ for some $n$, what would you try to multiply the right and left hand side with, so that you get $(n+1)^3$ and $2^{n+1}$ respectively? Compare those terms. — Sarvesh Ravichandran Iyer, Dec 11 '17 at 05:29
See https://math.stackexchange.com/questions/1140396/prove-by-induction-that-n2n/1140398#1140398 — lab bhattacharjee, Dec 11 '17 at 05:31

score 1 · Accepted Answer · answered Dec 11 '17 at 05:40

Hint:

We have to show that $(k+1)^3<2^{k+1}$. Note that:

$$(k+1)^3=k^3\cdot\frac{(k+1)^3}{k^3}=k^3\left(\frac{k+1}k\right)^3\;.$$

Since $k\ge 10$, $\frac{k+1}k=1+\frac1k\le 1+\frac1{10}=\frac{11}{10}$, and therefore

$$\left(\frac{k+1}k\right)^3\le \,\, ??$$

score 0 · Answer 2 · answered Dec 11 '17 at 05:43

Note that $(k+1)^3+(k-1)^3 = 2k^3 + 6k$, so as long as $(k-1)^3 \ge 6k$, we'll have that $(k+1)^3 \le 2k^3$. This is equivalent to showing that $(k-1)^3 \ge 6(k-1)+6$. Now $$(k-1)^3=(k-1)^2(k-1)\ge 81(k-1) = 6(k-1)+75(k-1)\ge 6(k-1)+6.$$ Hence the inequality holds, which implies that $(k+1)^3\le 2k^3$.

Now we can do induction. Certainly $10^3\le 2^{10}$ as you wrote. Then if $n^3 \le 2^n$, we have $$(n+1)^3 \le 2n^3 \le 2(2^n)=2^{n+1}.$$ Hence the induction holds.

Robin Grajeda · Answer 3 · 2017-12-11T08:11:12.210

Define $$S=\left\{ n\in %TCIMACRO{\U{2115} }% %BeginExpansion \mathbb{N} %EndExpansion ,\text{ }n\geq 10\text{ }|\text{ }n^{3}\leq 2^{n}\right\} $$

For $n=10$

$$ 10^{3}\leq 2^{10} $$

Therefore $S\neq \varnothing .$

Suppose that $k^{3}\leq 2^{k}$. We want to show that for any positive intger $k$, $\left( k+1\right) \in S.$ In other words $$\left( k+1\right) ^{3}\leq 2^{k+1}$$

Using the inductive hypothesis, we may write

$$ 2k^{3}\leq 2\left( 2^{k}\right) =2^{k+1} $$

All that remains to be shown is that $\left( k+1\right) ^{3}\leq 2k^{3}$ for $k\geq 10.$ This is certainly true for $k>3,$ so the requirement is satisfied.

Therefore, $S\subset %TCIMACRO{\U{2115} }% %BeginExpansion \mathbb{N} %EndExpansion .$

Holo · Answer 4 · 2017-12-11T15:08:46.233

When you have $$k^3<2^k$$ you can do 2 things, one multiply by 2 and get $2k^3<2^{k+1}$ and prove that $(k+1)^3<2k^3$ for $k\ge 10$ or you can multiply both sides by $\frac{(k+1)^3}{k^3}$ and get $(k+1)^3<\frac{(k+1)^3}{k^3}2^k$ then you prove that $\frac{(k+1)^3}{k^3}<2$ for $k\ge 10$

I'll show the later:

You have $\left(1+\frac1k\right)^3\le 1.1^3<2$

Hence $$(k+1)^3<\frac{(k+1)^3}{k^3}2^k<2\cdot2^k=2^{k+1}$$

Proof for Induction Question

4 Answers4