- Data Structure
- Networking
- RDBMS
- Operating System
- Java
- MS Excel
- iOS
- HTML
- CSS
- Android
- Python
- C Programming
- C++
- C#
- MongoDB
- MySQL
- Javascript
- PHP
- Physics
- Chemistry
- Biology
- Mathematics
- English
- Economics
- Psychology
- Social Studies
- Fashion Studies
- Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
For any positive integer n, prove that $n^3-n$ is divisible by 6.
Given:
$n^3\ -\ n$
To prove:
Here we have to prove that for any positive integer n, prove that $n^3-n$ is divisible by 6.
Solution:
Let us consider that:
$x\ =\ n^3\ –\ n$
Taking n common:
$x\ =\ n(n^2\ –\ 1)$
$x\ =\ n(n^2\ –\ 1^2)$
Using property {$a^2-b^2=$ (a $+$ b)(a $-$ b)}:
$x\ =\ n(n\ +\ 1)(n\ –\ 1)$
We know that (n $-$ 1), (n) and (n $+$ 1) are three consecutive numbers. So we can conclude that:
- One number must be even, and $x$ is divisible by 2.
- One number must be multiple of 3, and $x$ is divisible by 3 also.
Now,
If a number is divisible by 2 and 3 both then that number is divisible by 6.
So, $n^3\ –\ n$ is divisible by 6 for any positive integer n.
Advertisements