Question

A triangular number which is the sum of the square of two consecutive odd numbers is?

• A. $10$
• B. $15$
• C. $21$
• D. $28$

Answer 1

The correct option is A $10$

Let the two consecutive odd numbers be $n-2,n$

$(n-2{)}^2+n{n}^2+4-4n+n^{2}2n^2+4-4n$

If we put $n=12\left(1\right)-4+4=2n=22\left(4\right)+4-4=4n=32\left(9\right)+4-4\left(3\right)=18+4-12=10$

$\therefore$ Triangular no. is $10$

For any other value of $n$ the triangular no. is not in the required options

$\therefore$ we take $n=3$ and triangular no. is $10$