Question

What is the ninth term in the following sequence of natural numbers: 1, 1+3, 1+3+5, 1+3+5+7, .... ?

• A. 81
• B. 64
• C. 100
• D. 49

Answer 1

The correct option is A 81

First term $=1=1^2$
Second term $=1+3=4=2^2$
Third term $=1+3+5=9=3^2$
Fourth term $=1+3+5+7=16=4^2$
and so on.
We can observe a general pattern of the nth term being $n^2$.
Example : Fifth term $=5^2$

Ninth term $=9^2=81$
The sum of first odd natural numbers is always a perfect square.