Sara makes a staircase out of toothpicks as shown:

Hi,

n=3 -> we require 4 TP for the first, 3 for the 2 below, and 8 for the third row. Realize that for the third row, we need 3 for the first and last stair, but only 2 for the second! This pattern continues for n=4, where again, in the 4th row, the first and last stair require 3 TP, but the 2 in the middle only require two. We see that each row after the first one consists of staircases which either require 3 or 2 TP.
We can come up with a formula. We always need 4 for the first, we have (n-1)*2 staircases which require 3 TP (to see this, try values for n from 2 to 5 and you will see the pattern) -> (n-1)*2= Total Number of 3TP staircases

For the total number of 2 TP staircases, we can use a modified version of the sum formula. Why? Here is the pattern as to how many staircases we have which require 2 TP:

n=2 -> 0
n=3 -> 1
n=4 -> 2
n=5 -> 3
...

The above means that row 3 has 1 2 TP staircase, row 4 has 2 2TP staircase, and so on. But we have to add them all up. For n=5, we have a total of 1+2+3=6 2TP staircases. Sum formula is:
(n*(n+1))/2. But since we only start at n=3 (before, it is 0 2TP staircases), we modify it to:

((n-2)*(n-1))/2=Number of 2TP staircases

Combining all the info above: We know that the first stair needs 4, and afterwards, we have the formula for how many 2TP staircases and how many 3TP staircases we have, giving us in total:

Total Number of TP=4+(n-1)*2*3+(((n-2)*(n-1))/2)*2, plugging in the numbers we get
180=4+(n-1)*6+(n-2)*(n-1)
182=6n+n^2-3n+2
180=3n+n^2
180=n(3+n)

Plugging in numbers, we see that for n=12 we get
180=12*15 -> (C)