# Inequality proof

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

Note by Abdelfatah Teamah
1 year, 5 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

• bulleted
• list

1. numbered
2. 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 1

paragraph 2

paragraph 1

paragraph 2

> 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:

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.

- 1 year, 5 months ago