Inequality proof

Prove that if $n>2$ , then $2^{n+1} > n^2 + n + 2$ holds true.

#Algebra

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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

Proof By Induction:

Base Case: $n=3$

$2^{3+1}>3²+3+2\implies 16>14$ ,which is true.

Inductive Step: Let the statement be true for some $k$ .We will prove that it holds for $k+1$ as well.

By the inductive hypothesis we have $2^{k+1}>k²+k+2$

We have to prove that $2^{k+2}>(k+1)²+(k+1)+2$

Proof: $2^{k+2}=2(2^{k+1})>2(k²+k+2)>(k+1)²+(k+1)+2$ The last inequality, after simplifying becomes $k²>k$ ,which is true as $k>2$ .

Hence Proved.

Abdur Rehman Zahid - 4 years, 4 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}`	$\boxed{123}$