I'm modelling a system using a Markov chain and by a few iterations of the transition matrix I can see a pattern emerging in the resulting polynomial that really looks like Pascal's triangle, but isn't quite. I've been out of the mathematics world for a little while now so I may be missing quite obvious, but this feels like the sort of thing that should be able to wrap very neatly up into a little summation or something. Ideally I want a simple expression in terms of $n$ and $\mu$.
$$T=\begin{bmatrix} \mu^2-2\mu+1 & \mu-\mu^2 & \mu-\mu^2 & \mu^2\\ 0 & 1-\mu & 0 & \mu\\ 0 & 0 & 1-\mu & \mu\\ 0 & 0 & 0 & 1 \end{bmatrix}$$
$$i=\begin{bmatrix} 1\\ 0\\ 0\\ 0 \end{bmatrix}$$
Basically what I've done so far is take my transition matrix $T$ and raise it to the powers 1-4, then left multiplying by the initial state vector $i$. The first term of the resulting state vector $p$ is pretty clearly $(\mu^2-2\mu+1)^n$, but the other ones are much bigger. At the moment I'm only interested in the final term $p_4$, so I'll only show that one.
Here's the polynomials in expanded form: $$\mu^2\\ \mu^4-4\mu^3+4\mu^2\\ \mu^6-6\mu^5+15\mu^4-18\mu^3+9\mu^2\\ \mu^8-8\mu^7+28\mu^6-56\mu^5+68\mu^4-48\mu^3+16\mu^2$$
There's clearly a pattern here, the powers are going up by two each time, and each coefficient alternates sign, but what I'm trying to figure out is why the last few coefficients of each polynomial don't seem to follow the Pascal's triangle pattern. In each equation the middle coefficient is 2 smaller than the binomial coefficient should be, and then the ones after that are different in a way I can't seem to find a pattern for.
I was hoping someone might be able to point me in the right direction on this, I feel like there must be a way to write a general expression for this but I can't put my finger on it.
A standard technique for evaluating arbitrary powers $n$ of a square matrix $T$ is to use the Jordan decomposition $$T = QJQ^{-1} ,$$ where $J$ is the Jordan normal form of $T$ (its diagonal entries are the eigenvalues of $T$) and $Q$ is the corresponding similarity matrix. Then, $$T^n = (Q J Q^{-1})^n = Q J^n Q^{-1} ,$$ and it's comparatively easy to compute the $n$th power $J^n$ of a Jordan matrix.
In our case, in terms of $$\nu := 1 - \mu ,$$ we have $$J = \operatorname{diag}(\nu^2, \nu, \nu, 1), \qquad \qquad Q = \pmatrix{ \frac{1}{\nu} & -\frac{\nu + 1}{\nu} & \cdot & 1 \\ \cdot & \frac{\nu^2 - \nu - 1}{(\nu + 1) \nu} & \frac{\nu}{\nu + 1} & 1 \\ \cdot & -\frac{2 \nu + 1}{\nu + 1} & -\frac{\nu}{\nu + 1} & 1 \\ \cdot & \cdot & \cdot & 1} .$$ Our $J$ is diagonal, and it's easy to compute the $n$th power of a diagonal matrix---you simply take the $n$th power of each entry on the diagonal, so $J^n = \operatorname{diag}(\nu^{2n}, \nu^n, \nu^n, 1)$. Computing directly gives that the relevant entries of $T^n$ are $$\color{#bf0000}{ \boxed{ \begin{align} (T^n)_{11} &= \nu^{2n} \\ (T^n)_{12} = (T^n)_{13} &= \nu^n (1 - \nu^n) \\ (T^n)_{14} &= (1 - \nu^n)^2 \end{align}}} ,$$ which agrees with your computations for $n = 1, 2, 3, 4$.
Remark Expanding the square in the expression for $(T^n)_{14}$ gives $$(T^n)_{14} = \nu^{2 n} - 2 \nu^n + 1 = (1 - \mu)^{2 n} + \textrm{(a polynomial of degree $n$ in $\mu$)} ,$$ which explains why the coefficients of the polynomials $(T^n)_{14}$ agree (up to sign) with those of the binomial coefficients initially (more precisely, for the terms of degree $> n$).