Help needed

Please help me with this problem based on Induction.

Prove the following:

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

#NumberTheory #Induction

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Math	Appears 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}`	$\boxed{123}$

Comments

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 ,

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

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

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

Nihar Mahajan - 5 years, 8 months ago

@Nihar Mahajan

How is one supposed to think so beautifully?

Swapnil Das - 5 years, 8 months ago

Thanks! :)

Nihar Mahajan - 5 years, 8 months ago

Nihar really Impressive solution

Atul Shivam - 5 years, 8 months ago

Thanks! :)

Nihar Mahajan - 5 years, 8 months ago

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

Calvin Lin Staff - 5 years, 8 months ago

Math	Appears 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}`	$\boxed{123}$