Find the sum of first n odd natural numbers.
(a) $n^3$
(b) $n(n + 1)$
(c) $n(n + 1)^2$
(d) $n^2$

Hint: First write down all the numbers which are asked to add up. Then see if there is some sequence in the numbers or not.

Complete step-by-step answer:
1 + 3 + 5 + 7 + 9 + ∙∙∙ + m where m is the nth odd number.
Observe that the difference between two consecutive terms is 2 and constant throughout the sequence.
Apply the formula of sum of Arithmetic Progression. The formula says that
If there is a sequence with first term a and every next term is the sum of its previous term and a constant number d. So the sequence is
a, a + d, a + 2d, a + 3d, ∙∙∙
This sequence is called Arithmetic Progression.
Sum S of n terms of this sequence is determined by
$S=\dfrac{n}{2}\left[ a+\left( n-1 \right)d \right]$

We have to find the sum of first n odd natural numbers. So let us first down all the numbers which are to be added.
1 + 3 + 5 + 7 + 9 + ∙∙∙ + m , where m is the nth odd natural number.
We see that the difference between two consecutive terms is always 2. That means the difference between consecutive terms is constant. Hence this series is an Arithmetic Progression.
First term of this sequence is 1 and the common difference is 2.
We know the formula of arithmetic progression as
If a, a + d, a + 2d, a + 3d, ∙∙∙ is a arithmetic sequence then the sum S of n terms of this sequence is given by
$S=\dfrac{n}{2}\left[ a+\left( n-1 \right)d \right]$
So we get our sequence is arithmetic progression with a = 1 and d = 2.
So the sum of first n term of this sequence is given by
$\begin{align}
  & S=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right] \\
 & \Rightarrow S=\dfrac{n}{2}\left[ 2\left( 1 \right)+\left( n-1 \right)2 \right] \\
 & \Rightarrow S=\dfrac{n}{2}\left[ 2n \right] \\
 & \Rightarrow S={{n}^{2}} \\
\end{align}$
Hence the sum of the first n odd natural number is $n^2$.
So, option (d) is correct.

Note: This question can be solved easily if we can write nth of the sequence in terms of n.
The 1st term is 1.
The 2nd term is 1 + 2 = 3.
The 3rd term is 1+2(2) = 5.
In this way, the nth term of this sequence can be written as 1 + (n - 1)2 = 2n – 1.
Now there is another formula for the sum of n terms.
 $S=\dfrac{n}{2}\left[ a+l \right]$
Where a is the first term and l is the nth term.
Using this we get
$\begin{align}
  & S=\dfrac{n}{2}\left[ 1+2n-1 \right] \\
 & \Rightarrow S=\dfrac{n}{2}\left[ 2n \right] \\
 & \Rightarrow S={{n}^{2}} \\
\end{align}$