Question: Define 
\begin{equation}
A = \begin{pmatrix} \frac{1}{2} &\frac{1}{2} & 0\\ 0& \frac{3}{4} & \frac{1}{4}\\ 0& \frac{1}{4} & \frac{3}{4} \end{pmatrix}.
\end{equation}

Note that the sum of the dimensions of the eigenspaces of $A$ is only two. $A$ is thus not diagonalizable. How can we compute $A^n$?

Answer: 
Here is a different way using a rather classical trick, converting the issue into a binomial expansion. Indeed, we can write :

$$A=\frac12(I+B) \ \text{where} \ B:=\begin{pmatrix}0&1&0 \\0&1/2&1/2\\0&1/2&1/2\end{pmatrix}$$

where matrix $B$ has the following particularity

$$B^n=C \ \text{for all} \ n>1 \ \text{where} \ C:=\begin{pmatrix}0&1/2&1/2\\0&1/2&1/2\\0&1/2&1/2\end{pmatrix}$$

Therefore 

$$A^n = \dfrac{1}{2^n}\left(I+\binom{n}{1}B+\binom{n}{2}B^2+\binom{n}{3}B^3+\cdots+\binom{n}{n}B^n\right)$$

$$A^n = \dfrac{1}{2^n}\left(I+nB+\binom{n}{2}C+\binom{n}{3}C+\cdots+\binom{n}{n}C\right)\tag{1}$$

As is well known, $\sum\_{k=0}^n \binom{n}{k}=2^n$, reducing (1) to :

$$A^n = \dfrac{1}{2^n}\left(I+nB+(2^n-n-1)C\right)$$

It suffices now to replace $B$ and $C$ by their expression 

$$A^n = \dfrac{1}{2^n}\left(\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}+n\begin{pmatrix}0&1&0 \\0&1/2&1/2\\0&1/2&1/2\end{pmatrix}+(2^n-n-1)\begin{pmatrix}0&1/2&1/2\\0&1/2&1/2\\0&1/2&1/2\end{pmatrix}\right)$$

to get the result (coinciding with yours !).
Note that your matrix $A$ has the generalized eigenvectors

\begin{equation}v\_1=\begin{pmatrix}1 \\ 0 \\ 0\end{pmatrix}, v\_2=\begin{pmatrix}0 \\ 2 \\ -2\end{pmatrix}, v\_3=\begin{pmatrix}0\\ 0 \\ 1\end{pmatrix}.\end{equation}

Thus, by Jordan decomposition, $A=\big(v\_1,v\_2,v\_3\big)J\big(v\_1,v\_2,v\_3\big)^{-1}$, where 

\begin{equation}J=\begin{pmatrix}\frac12 & 1 & 0\\0 & \frac12 & 0\\0 & 0 & 1\end{pmatrix}.\end{equation}

The problem of calculating $A^n$ is thus reduced to calculating $J^n$. Let $a\_{ij}^{(n)}$ denote the entry of $J^n$ in the $i$-th row and $j$-th column. 

The product of an arbitrary $3\times3$-matrix with $J$ is given by:
\begin{equation}
\begin{pmatrix}
a&b&c\\d&e&f\\g&h&i
\end{pmatrix} J =
\begin{pmatrix}
\frac a2&a+\frac b2&c\\\frac d2&d+\frac e2&f\\\frac g2&g+\frac h2&i
\end{pmatrix}.
\end{equation}

We can deduce that, for all $n\in\Bbb N$: 
\begin{align}
a\_{11}^{(n)}&=a\_{22}^{(n)}=\frac1{2^n}, \\a\_{21}^{(n)}&=a\_{31}^{(n)}=0,\\ 
a\_{13}^{(n)}&=a\_{23}^{(n)}=a\_{32}^{(n)}=0, \\
a\_{33}^{(n)}&=1,\\
a\_{12}^{(n+1)}&=a\_{11}^{(n)}+\frac{a\_{12}^{(n)}}2=\frac1{2^n}+\frac{a\_{12}^{(n)}}2.
\end{align}

Thus, all $a\_{ij}^{(n)}$ are explicitly known except for $a\_{12}^{(n)}$. Note that, by the last equation, \begin{equation}a\_{12}^{(n+1)}=2^{-n}+\frac{a\_{12}^{(n)}}2 = 2^{-n}+2^{-n}+\frac{a\_{12}^{(n-1)}}4 = \dots = (n+1)\cdot2^{-n}.\end{equation}

Thus, 
\begin{equation}J^n=\begin{pmatrix}2^{-n}&n\cdot 2^{1-n} & 0\\0 & 2^{-n} & 0\\0 & 0 & 1\end{pmatrix}.\end{equation}

And by some calculations, we find that 
\begin{equation}
A^n=\big(v\_1,v\_2,v\_3\big)J^n\big(v\_1,v\_2,v\_3\big)^{-1}=
\begin{pmatrix}
2^{-n} & n\cdot 2^{-n-1} - 2^{-n-1} + \frac12 & {1-\frac{n+1}{2^n}\over2}\\
0 & {2^{-n}+1\over2} & {1-2^{-n}\over2} \\
0 & {1-2^{-n}\over2} & {2^{-n}+1\over2} 
\end{pmatrix}.
\end{equation}