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]
1. **Understanding the Problem:**
   - We have a tetrahedral stack of spheres.
   - Each row \( n \) has a triangular number of spheres, given by \( T_n = \frac{n(n+1)}{2} \).
   - The number assigned to each sphere is the sum of the numbers assigned to the spheres directly above it in the previous row.
   - We need to find the sum of the numbers assigned to each sphere from row \( 0 \) to row \( n \).

2. **Analyzing the Pattern:**
   - The top sphere in row \( 0 \) is assigned the number \( 1 \).
   - For row \( 1 \), each of the 3 spheres is assigned the number \( 1 \) (since they all touch the top sphere), so the sum is \( 3 \).
   - For row \( 2 \), each sphere is assigned the sum of the numbers of the spheres it touches in row \( 1 \). Since each sphere in row \( 1 \) is \( 1 \), and each sphere in row \( 2 \) touches 3 spheres in row \( 1 \), the sum is \( 3^2 = 9 \).
   - This pattern continues such that the sum of the numbers in row \( n \) is \( 3^n \).

3. **Summing the Series:**
   - We need to find the sum of the numbers from row \( 0 \) to row \( n \):
     \[
     S = 1 + 3 + 3^2 + 3^3 + \cdots + 3^n
     \]
   - This is a geometric series with the first term \( a = 1 \) and common ratio \( r = 3 \).

4. **Sum of a Geometric Series:**
   - The sum of the first \( n+1 \) terms of a geometric series is given by:
     \[
     S = a \frac{r^{n+1} - 1}{r - 1}
     \]
   - Substituting \( a = 1 \) and \( r = 3 \):
     \[
     S = \frac{3^{n+1} - 1}{3 - 1} = \frac{3^{n+1} - 1}{2}
     \]

Conclusion:
\[
\boxed{\frac{3^{n+1} - 1}{2}}
\]