In the pattern of numbers shown, every row begins with a 1 and ends with a 2. Each of the numbers, not on the end of a row, is the sum of the two numbers located immediately above and to the right, and immediately above and to the left. For example, in the fourth row the 9 is the sum of the 4 and the 5 in the third row. If this pattern continues, the sum of all of the numbers in the thirteenth

![](https://cdn.mathpix.com/cropped/2024_04_20_46ead6524a8d61e21c51g-086.jpg?height=285&width=293&top_left_y=1630&top_left_x=1404)
row is
(A) 12270
(B) 12276
(C) 12282
(D) 12288
(E) 12294
To get a better understanding of the pattern, let us write each of the numbers in the way that it is obtained:

$$
\begin{aligned}
& \begin{array}{llll}
1 & 2 & 1 & 2
\end{array} \\
& \begin{array}{llllll}
1 & 3 & 2 & 1 & 1+2 & 2
\end{array}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-369.jpg?height=46&width=637&top_left_y=687&top_left_x=514)

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-369.jpg?height=48&width=735&top_left_y=735&top_left_x=478)

$$
\begin{aligned}
& \therefore \quad \begin{array}{lllll}
\therefore & \ddots & \therefore & \vdots & \ddots
\end{array}
\end{aligned}
$$

We can see from these two patterns side by side that each number in a row is accounted for twice in the row below. (The 1 or 2 on the end of a row appears again at the end of the row and as part of the sum in one number in the next row. A number in the middle of a row appears as part of the sum in two numbers in the next row.)

Therefore, the sum of the numbers in a row should be two times the sum of the numbers in the previous row.

We can check this:

Sum of the numbers in the 1st row 3

Sum of the numbers in the 2nd row 6

Sum of the numbers in the 3rd row $\quad 12$

Sum of the numbers in the 4 th row $\quad 24$

Thus, the sum of the numbers in the thirteenth row should be the sum of the elements in the first row multiplied by 2 twelve times, or $3 \times 2^{12}=3 \times 4096=12288$.

ANsWER: (D)