Induction is done by demonstrating that if the condition is true for some $n$ then it must also be true for $n+1$. If you then show that the condition is true for $n=0$ then it must be true for all $n>0$. For this problem:
Step $1$: $n=1$
The sum of the first $1$ odd numbers is $1$. $1^2=1$. Therefore the condition holds for $n=1$.
Step $2$: induction
If the sum of the first $n$ odd numbers is $n^2$ then the sum of the first $n+1$ integers is
$n^2 + (2n + 1) = (n+1)(n+1)=(n+1)^2$
So the condition is also true for $n+1$.
Step $3$: conclusion
Since the we have shown that the condition is true for $n=1$ and we have shown that if it is true for $n$ then it is also true for $n+1$ then it follows by induction that it is true for all $n\geq 1$.
The first step is
to get this into
mathematical form.
I would write it like this:
Let
$(a_i)_{i=1}^n$
be odd integers.
Then,
for any positive integer $m$,
$\sum_{i=1}^{2m} a_i$
is even
and
$\sum_{i=1}^{2m-1} a_i$
is odd.
Proof.
Note:
All variables are integers.
The basic facts needed are that
(1) every even number $a$
can be written in the form
$a = 2b$;
(2) every odd number $a$
can be written in the form
$a = 2b+1$;
(3) all numbers are either
even or odd;
(4) the sum of two even numbers
is even;
(5) the sum of an odd and even integer
is odd;
(6) the sum of two odd numbers
is even.
Note:
Facts (1) and (2)
are definitions.
A good exercise is
to prove facts
(3) through (6).
For $m=1$,
this states that
$a_1$ is odd and
$a_1+a_2$ is even.
The first is true by assumption.
The second is true because
the sum of two odd integers
is even.
For the induction step,
suppose it is true for $m$.
The statement for
$m+1$ is
$\sum_{i=1}^{2(m+1)} a_i$
is even
and
$\sum_{i=1}^{2(m+1)-1} a_i$
is odd.
For the first,
$\begin{array}\\
\sum_{i=1}^{2(m+1)} a_i
&=\sum_{i=1}^{2m+2} a_i\\
&=\sum_{i=1}^{2m} a_i+a_{2m+1}+a_{2m+2}\\
&=(\sum_{i=1}^{2m} a_i)+(a_{2m+1}+a_{2m+2})\\
\end{array}
$
and this is even
(using fact 4) because
$\sum_{i=1}^{2m} a_i$
is even by the induction hypothesis
and
$a_{2m+1}+a_{2m+2}$
is even by fact 6.
For the second,
$\begin{array}\\
\sum_{i=1}^{2(m+1)-1} a_i
&=\sum_{i=1}^{2m+1} a_i\\
&=\sum_{i=1}^{2m-1} a_i+a_{2m}+a_{2m+1}\\
&=(\sum_{i=1}^{2m-1} a_i)+(a_{2m}+a_{2m+1})\\
\end{array}
$
and this is odd
(using fact 5) because
$\sum_{i=1}^{2m-1} a_i$
is odd by the induction hypothesis
and
$a_{2m}+a_{2m+1}$
is even by fact 6.
You could also group the sum as
$\sum_{i=1}^{2m} a_i+a_{2m+1}
$;
in this,
the sum is even
and $a_{2m+1}$
is odd,
so their sum is odd.
Best Answer
There is a simple rule: For any arithmetic progression, that is any series where the difference between consecutive elements is constant, the sum is equal to the number of elements, multiplied by the average between the first and the last element.
In your case, the first element is 1, the last element is 2n - 1, the average is n, there are n elements, therefore the sum is $n^2$.