How much weight is on each person in a human pyramid? After participating in a human pyramid and learning that it's very uncomfortable to have a lot of weight on your back, I figured I'd try to write out a recurrence relation for the total amount of weight on each person in the pyramid so that I could see just how much weight I ended up carrying.
There are many different kinds of human pyramids, but the sort of pyramid I'm referring to is one that looks like this:
               *
              / \
             *   *
            / \ / \
           *   *   *
          / \ / \ / \
         *   *   *   *
               ...

Here, each star is a person and the lines represent how each person is supported.
I'm going to make the unrealistic assumption that each person weighs the same amount, which I'll call $W$.  I'm also going to assume that each person evenly transmits their weight (plus the total weight on top of them) to the two people below them.
Given these assumptions, I came up with a recurrence $w_{i, j}$ that says how much weight the $i$th person in row $j$ of the human pyramid carries on their back.  It ended up coming out like this:


*

*$w_{1, 1} = 0$.  The top person in the pyramid has no weight on them.

*$w_{1, n+1} = \frac{w_{1, n} + W}{2}$  The very first person on each row is shouldering half the weight of the person above them.  That weight is given by the weight of the person ($W$) plus the load that person carries ($w_{1, n}$).

*$w_{n+1, n+1} = \frac{w_{n, n} + W}{2}$.  The last person on each row shoulders half of the weight of the person above them.

*$w_{k+1, n+1} = \frac{w_{k+1, n} + w_{k, n} + 2W}{2}$.  Each person on a row other than the first or the last shoulders half the weight from each of the people above them.  The two people above them have $w_{k+1, n}$ and $w_{k, n}$ weight on them, and each one independently weighs $W$.  Half of each of these weights is transmitted to the person below.


I was able to write a computer program that evaluated this recurrence and I was able to get values from it, but I have no idea how to find a closed-form expression for this recurrence.  It's somewhat similar to the recurrence for combinations - each term is expressed as a sum of the two terms above it - but there's some extra junk thrown in as well.
Is there a standard approach for simplifying recurrences like this?
Thanks!

For a generating function approach, Let the coefficient of $x^k$ in $P_n(x)$ be the weight on the shoulders of person $k$ (starting at $k=0$ on the left) in row $n$ (starting at $n=0$ at the top). To compute $P_n(x)$ from $P_{n-1}(x)$, we first add the weight of the people in row $n-1$ to the weights on their shoulders:
$$
P_{n-1}(x)+\frac{1-x^n}{1-x}\tag{1}
$$
Those people distribute half their weight to each person below them, so we get
$$
P_n(x)=\frac{1+x}2\left(P_{n-1}(x)+\frac{1-x^n}{1-x}\right)\tag{2}
$$
Solving this recurrence yields
$$
P_n(x)=\frac{1+x}{(1-x)^2}\left(1-2\left(\frac{1+x}{2}\right)^{n+1}+x^{n+1}\right)\tag{3}
$$
$(3)$ can be checked by noting that $P_0(x)=0$ and then by checking that $(3)$ satisfies $(2)$.
Using the binomial theorem to multiply the terms in $(3)$, we get that $w_{m+1,n+1}$, the coefficient of $x^m$ in $P_n(x)$, is given by
$$
2m+1-\frac1{2^n}\sum_{k=0}^m(k+1)\binom{n+2}{m-k}\tag{4}
$$
There might be some special function of which I am unaware (a hypergeometric function perhaps), but in terms of common functions, $(4)$ is as close to a closed form as I think there is.

Solving the recurrence
Let $Q_n(x)=\left(\frac2{1+x}\right)^nP_n(x)$, then multiplying $(2)$ by $\left(\frac2{1+x}\right)^n$ yields
$$
\begin{align}
Q_n(x)
&=Q_{n-1}(x)+\frac{1-x^n}{1-x}\left(\frac2{1+x}\right)^{n-1}\\
&=Q_{n-1}(x)+\frac12\frac{1+x}{1-x}\left(1-x^n\right)\left(\frac2{1+x}\right)^n\\
&=Q_{n-1}(x)+\frac12\frac{1+x}{1-x}\left(\left(\frac2{1+x}\right)^n-\left(\frac{2x}{1+x}\right)^n\right)\tag{5}
\end{align}
$$
Summing the geometric series gets a bit messy,
$$
\begin{align}
Q_n(x)
&=\frac12\frac{1+x}{1-x}\sum_{k=1}^n\left(\left(\frac2{1+x}\right)^k-\left(\frac{2x}{1+x}\right)^k\right)\\
&=\frac12\frac{1+x}{1-x}\left(\frac2{1+x}\frac{\left(\frac2{1+x}\right)^n-1}{\frac2{1+x}-1}-\frac{2x}{1+x}\frac{\left(\frac{2x}{1+x}\right)^n-1}{\frac{2x}{1+x}-1}\right)\\
&=\frac{1+x}{1-x}\left(\frac{\left(\frac2{1+x}\right)^n-1}{1-x}+x\frac{\left(\frac{2x}{1+x}\right)^n-1}{1-x}\right)\\
&=\frac{1+x}{(1-x)^2}\left(\left(\frac2{1+x}\right)^n-1+x\left(\frac{2x}{1+x}\right)^n-x\right)\\
&=\frac{1+x}{(1-x)^2}\left(\left(\frac2{1+x}\right)^n\left(1+x^{n+1}\right)-(1+x)\right)\tag{6}
\end{align}
$$
Multiplying by $\left(\frac{1+x}2\right)^n$, we arrive at $(3)$:
$$
P_n(x)=\frac{1+x}{(1-x)^2}\left(1+x^{n+1}-2\left(\frac{1+x}2\right)^{n+1}\right)\tag{7}
$$

Examples:
$$
\begin{align}
P_0(x)&=0\\
P_1(x)&=\frac12+\frac12x\\
P_2(x)&=\frac34+\frac32x+\frac34x^2\\
P_3(x)&=\frac78+\frac{17}{8}x+\frac{17}{8}x^2+\frac78x^3\\
P_4(x)&=\frac{15}{16}+\frac52x+\frac{25}{8}x^2+\frac52x^3+\frac{15}{16}x^4
\end{align}
$$
The bottom-middle person on a $5$-tier pyramid is supporting more than $3$ people's weight on their back.
