Waste less time on Facebook — follow Brilliant.
×

Mathematical induction proof?

what is the most simplest method to solve this question by using mathematical induction to prove that \(n !>2^n-1 \) and integers \(n>5\)?

Note by Syed Hissaan
1 year, 2 months 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

@Calvin Lin @Chung Kevin , any proof for that one !

Syed Hissaan - 1 year, 2 months ago

Log in to reply

What have you tried? What have you done?

Seems to me like a simple induction proof starting with base case of \( n = 6 \).

Calvin Lin Staff - 1 year, 2 months ago

Log in to reply

is it valid in this type of proof to place any value of the k < 6 on one of the side to setup our equations)# for n=k e.g : \( (k+1)!> 2^k .2 -1 +2^k -k \) can we take k<4 * or *k=2on the R.H.S to proof our question

Syed Hissaan - 1 year, 2 months ago

Log in to reply

@Syed Hissaan No, because the statement is not true for \( k \leq 4 \). That is why you have to start the base case in the appropriate domain of \( n > 5 \) (or whatever they tell you to).

Calvin Lin Staff - 1 year, 2 months ago

Log in to reply

@Calvin Lin ok thanks !

Syed Hissaan - 1 year, 2 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...