Waste less time on Facebook — follow Brilliant.
×

Inequality proof

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

Note by Abdelfatah Teamah
9 months, 2 weeks ago

No vote yet
1 vote

  Easy Math Editor

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

[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} \( \boxed{123} \)

Comments

Sort by:

Top Newest

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 - 9 months, 2 weeks ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...