The sum of the squares of 3 consecutive positive integers is 116. What are the numbers?

Which of the following equations is used in the process of solving this problem?
3n2 + 5 = 116
3n2 + 3n + 3 = 116
3n2 + 6n + 5 = 116

Question

The sum of the squares of 3 consecutive positive integers is 116. What are the numbers?

Which of the following equations is used in the process of solving this problem?
3n2 + 5 = 116
3n2 + 3n + 3 = 116
3n2 + 6n + 5 = 116

anonymous · Answer

If you let the number as n, then other two consecutive numbers will be: n+1, n + 2

So:

$$n^2 + (n+1)^2 + (n+2)^2 = 116$$
Just solve this..

anonymous · Answer

To open up the squares you may use this:

$$(x+y)^2 = x^2 + y^2 + 2xy$$

anonymous · Answer

Idk how to do this

anonymous · Answer

Can you solve this:

$$(n + 1)^2$$

anonymous · Answer

Just show me how to slove it

anonymous · Answer

You just need to sustitute it in the formula (a+b)^2=a^2+2ab+b^2..can u do it now??.. Just take n as a and 1as b and substitute the equation @waterineyes has provided... Okay??