
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system
There is an infinitely tall tetrahedral stack of spheres where each row of the tetrahedron consists of a triangular arrangement of spheres, as shown below. There is  $1$  sphere in the top row (which we will call row  $0$ ),  $3$  spheres in row  $1$ ,  $6$  spheres in row  $2$ ,  $10$  spheres in row  $3$ , etc. The top-most sphere in row  $0$  is assigned the number  $1$ . We then assign each other sphere the sum of the number(s) assigned to the sphere(s) which touch it in the row directly above it. Find a simplified expression in terms of  $n$  for the sum of the numbers assigned to each sphere from row  $0$  to row  $n$ .

[asy]
import three;
import solids;
size(8cm);

//currentprojection = perspective(1, 1, 10);

triple backright = (-2, 0, 0), backleft = (-1, -sqrt(3), 0), backup = (-1, -sqrt(3) / 3, 2 * sqrt(6) / 3);

draw(shift(2 * backleft) * surface(sphere(1,20)), white); //2
draw(shift(backleft + backright) * surface(sphere(1,20)), white); //2
draw(shift(2 * backright) * surface(sphere(1,20)), white); //3
draw(shift(backup + backleft) * surface(sphere(1,20)), white);
draw(shift(backup + backright) * surface(sphere(1,20)), white);
draw(shift(2 * backup) * surface(sphere(1,20)), white);

draw(shift(backleft) * surface(sphere(1,20)), white);
draw(shift(backright) * surface(sphere(1,20)), white);
draw(shift(backup) * surface(sphere(1,20)), white);

draw(surface(sphere(1,20)), white);

label("Row 0", 2 * backup, 15 * dir(20));
label("Row 1", backup, 25 * dir(20));
label("Row 2", O, 35 * dir(20));

dot(-backup);
dot(-7 * backup / 8);
dot(-6 * backup / 8);

dot((backleft - backup) + backleft * 2);
dot(5 * (backleft - backup) / 4 + backleft * 2);
dot(6 * (backleft - backup) / 4 + backleft * 2);

dot((backright - backup) + backright * 2);
dot(5 * (backright - backup) / 4 + backright * 2);
dot(6 * (backright - backup) / 4 + backright * 2);
[/asy]

Present the answer in LaTex format: \boxed{Your answer}
user