# Help needed

Prove the following:

$$2^{k} > k^{3}$$ , $$\forall$$ $$k>9$$

Note by Swapnil Das
2 years, 9 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

Induction base: $$2^{10}=1024>1000=10^3$$ . Thus the inequality holds for $$k=10$$.

Induction step: Lets assume that $$2^k > k^3 \forall k>9$$.

Thus , $$2^{k+1}=2\times 2^k > 2k^3$$.

Note that $$2k^3 = (k+1)^3+k^3-3k^2-3k-1$$. Hence ,

\begin{align} 2^{k+1} &> (k+1)^3+k^3-3k^2-3k-1 \\ &> (k+1)^3+k^3-3k^2-3k-459 \\ &= (k+1)^3+(k-9)(k^2+6k+51) \end{align}

Note that $$k^2+6k+51 > 0$$ since its discriminant is negative.

Thus finally we have $$2^{k+1}>(k+1)^3+(k-9)(0) = (k+1)^3 \forall \ k > 9$$.

- 2 years, 9 months ago

Nihar really Impressive solution

- 2 years, 9 months ago

Thanks! :)

- 2 years, 9 months ago

How is one supposed to think so beautifully?

- 2 years, 9 months ago

Thanks! :)

- 2 years, 9 months ago

Yup, definitely check out Induction. This is covered under the inequalities section.

Staff - 2 years, 9 months ago