Write the sum of first n odd natural numbers.

Hint: Natural numbers are the numbers that start from 1 and go till infinity.
Now, the formula for the sum of n natural numbers is
Sum \[={n\left( n+1 \right)}/{2}\] .

Complete step-by-step answer:

We know that, an odd number can represented as
n=2k+1 (where n is an odd number and k can be any natural number)
Now, an odd natural number is same as an odd number and can be represented as mentioned above as
n=2k+1
Now, the sum that we are evaluating is
S=1+3+5+7+9+………………………+(2k+1) (This is the sum of odd natural numbers)
Now, the sum of all odd natural numbers till \[{{n}^{th}}\] odd natural number, we need to just put ‘n’ in place ok ‘k’ and hence the sum that we will get would be equal to
S=1+3+5+7+9+………………………….+ (2n+1).
Now, this sum can be written as
S \[=\sum\limits_{r=1}^{n}{(2r+1)}\]
Now, the summation can be broken as follows
S \[=\sum\limits_{k=1}^{n}{2k}+\sum\limits_{k=1}^{n}{1}\]
Here, the first part is nothing but the same summation of first n natural numbers which is just multiplied by 2 and the second part is just adding 1 for n times.
So, the summation would look like as follows
\[\begin{align}
  & =2\times {n\left( n+1 \right)}{2}+n \\
 & =n(n+1)+n \\
 & ={{n}^{2}}+n+n \\
 & ={{n}^{2}}+2n \\
\end{align}\]
  Hence, the sum of first n odd natural numbers is \[{{n}^{2}}+2n\] .

NOTE: -
The students can make an error by manually counting first n odd natural numbers rather than using the summation formula for first n natural numbers can be used and the problem becomes very easy.