Refer to the attachment for the visualization of the staircase that is 4 units tall and contains 10 unit squares.
This problem can easily be solved by using the formula for arithmetic series. The formula is given by:
![S_{n}= \frac{n}{2}[2 a_{1}+(n-1)d]](https://tex.z-dn.net/?f=%20S_%7Bn%7D%3D%20%5Cfrac%7Bn%7D%7B2%7D%5B2%20a_%7B1%7D%2B%28n-1%29d%5D%20%20)
where S is the sum, n is the number of terms, 
is the first term, and d is the common difference.
In the case of the problem, n is 4, the first term is 1, and the common difference is also 1. If you substitute these conditions to the formula, you'll get 10 square units.
Now for the extended staircase, we will have 12 number of terms since the staircase is now 12 units tall. The common difference and the first term will still be the same as the previous condition. We can solve for the unit squares it contains by substituting these values to the formula for arithmetic series:
![S_{12}= \frac{12}{2}[2(1)+(12-1)(1)]](https://tex.z-dn.net/?f=%20S_%7B12%7D%3D%20%5Cfrac%7B12%7D%7B2%7D%5B2%281%29%2B%2812-1%29%281%29%5D%20)
![S_{12}= 6[2+11]=6(13)=78](https://tex.z-dn.net/?f=%20S_%7B12%7D%3D%206%5B2%2B11%5D%3D6%2813%29%3D78%20)
ANSWER: The extended staircase would contain 78 unit squares.
This problem can easily be solved by using the formula for arithmetic series. The formula is given by:
In the case of the problem, n is 4, the first term is 1, and the common difference is also 1. If you substitute these conditions to the formula, you'll get 10 square units.
Now for the extended staircase, we will have 12 number of terms since the staircase is now 12 units tall. The common difference and the first term will still be the same as the previous condition. We can solve for the unit squares it contains by substituting these values to the formula for arithmetic series:
ANSWER: The extended staircase would contain 78 unit squares.