The simplest prrof by induction involve finding a simple formula for the nth term of a sequence.
Proofs by induction have three parts.
1. Asume
is true. Often
or 1 so we are assuming
or![]()
could be the statement that the nth term of a sequence is some formula for that nth term – for example, the nth even number is
so
is the statement that the first even number is 2.
2. Assume
is true for some![]()
3. Prove
is true.
For the simple example above,
is true since the first even number is 2.
If
is true then the nth even number is![]()
Given an even number, to find the succeeding even number, add 2, so the (n+1)th even number is
so that
is true.
Example: Prove that the sum of the first
numbers is![]()
is the statement that the sum of the first 1 numbers is 1. Obviously this is true. Substitute
into (1) to give
so
is true.
Assume
is true for some
so that the sum of the first n numbers is![]()
The (n+1)th number is
We can add this to the sum of the first
numbers to get the sum of the first
numbers.

The statement
is the statement that the sum of the first
numbers is
so that
is true.