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Define
beginequation
A = beginpmatrix frac12 &frac12 & 0\ 0& frac34 & frac14\ 0& frac14 & frac34 endpmatrix.
endequation

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

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

$$A=frac12(I+B) textwhere B:=beginpmatrix0&1&0 \0&1/2&1/2\0&1/2&1/2endpmatrix$$

where matrix $B$ has the following particularity

$$B^n=C textfor all n>1 textwhere C:=beginpmatrix0&1/2&1/2\0&1/2&1/2\0&1/2&1/2endpmatrix$$

Therefore

$$A^n = dfrac12^nleft(I+binomn1B+binomn2B^2+binomn3B^3+cdots+binomnnB^nright)$$

$$A^n = dfrac12^nleft(I+nB+binomn2C+binomn3C+cdots+binomnnCright)tag1$$

As is well known, $sum_k=0^n binomnk=2^n$, reducing (1) to :

$$A^n = dfrac12^nleft(I+nB+(2^n-n-1)Cright)$$

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

$$A^n = dfrac12^nleft(beginpmatrix1&0&0\0&1&0\0&0&1endpmatrix+nbeginpmatrix0&1&0 \0&1/2&1/2\0&1/2&1/2endpmatrix+(2^n-n-1)beginpmatrix0&1/2&1/2\0&1/2&1/2\0&1/2&1/2endpmatrixright)$$

to get the result (coinciding with yours !).

Note that your matrix $A$ has the generalized eigenvectors

beginequationv_1=beginpmatrix1 \ 0 \ 0endpmatrix, v_2=beginpmatrix0 \ 2 \ -2endpmatrix, v_3=beginpmatrix0\ 0 \ 1endpmatrix.endequation

Thus, by Jordan decomposition, $A=big(v_1,v_2,v_3big)Jbig(v_1,v_2,v_3big)^-1$, where

beginequationJ=beginpmatrixfrac12 & 1 & 0\0 & frac12 & 0\0 & 0 & 1endpmatrix.endequation

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 $3times3$-matrix with $J$ is given by:
beginequation
beginpmatrix
a&b&c\d&e&f\g&h&i
endpmatrix J =
beginpmatrix
frac a2&a+frac b2&c\frac d2&d+frac e2&f\frac g2&g+frac h2&i
endpmatrix.
endequation

We can deduce that, for all $ninBbb N$:
beginalign
a_11^(n)&=a_22^(n)=frac12^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)+fraca_12^(n)2=frac12^n+fraca_12^(n)2.
endalign

Thus, all $a_ij^(n)$ are explicitly known except for $a_12^(n)$. Note that, by the last equation, beginequationa_12^(n+1)=2^-n+fraca_12^(n)2 = 2^-n+2^-n+fraca_12^(n-1)4 = dots = (n+1)cdot2^-n.endequation

Thus,
beginequationJ^n=beginpmatrix2^-n&ncdot 2^1-n & 0\0 & 2^-n & 0\0 & 0 & 1endpmatrix.endequation

And by some calculations, we find that
beginequation
A^n=big(v_1,v_2,v_3big)J^nbig(v_1,v_2,v_3big)^-1=
beginpmatrix
2^-n & ncdot 2^-n-1 - 2^-n-1 + frac12 & 1-fracn+12^nover2\
0 & 2^-n+1over2 & 1-2^-nover2 \
0 & 1-2^-nover2 & 2^-n+1over2
endpmatrix.
endequation