How to solve this recursion which is not homogenous I have the following recursion
$$a_n = \frac{1}{4}a_{n-1}+\frac{1}{4}(\frac{2}{3})^{n-1}$$
I've tried first to solve the homogeneous equation (shifting by one)
$$(E - \frac{1}{4})a_n = 0$$
where $Ea_n = a_{n+1}$ is the shift operator. The only solution to this equation is $E=\frac{1}{4}$. Now I thought that for a non-homogeneous equation, where the term $d(n)$ does not depend on the underlying recursion has the form $d(n) = k\mu^n$ and $\mu$ is not a root of the homogeneous equation, the solution is given by
$$a_n = \frac{k\mu^n}{\Phi(\mu)}$$
where $\Phi$ is the characteristic equation of the homogeneous one. In my case $d(n) = \frac{1}{4}\frac{2}{3}^{n}$, so $k=\frac{1}{4}$ and $\mu = \frac{2}{3}$. Thus the solution should be given by
$$a_n = \frac{\frac{1}{4}\frac{2}{3}^n}{\frac{2}{3}-\frac{1}{4}}=\frac{\frac{1}{4}\frac{2}{3}^n}{\frac{5}{12}}=\frac{3}{5}\frac{2}{3}^n$$
However, the solution should be $$\frac{3}{5}\frac{2}{3}^n-\frac{3}{5}\frac{1}{4}^n$$. What did I wrong?
Note: the question arises from another problem, see here

The telescoping summation helps: $$a_n=\frac{1}{4}a_{n-1}+\frac{1}{4}\left(\frac{2}{3}\right)^{n-1},$$
$$\frac{1}{4}a_{n-1}=\frac{1}{4^2}a_{n-2}+\frac{1}{4^2}\left(\frac{2}{3}\right)^{n-2},$$
$$\frac{1}{4^2}a_{n-2}=\frac{1}{4^3}a_{n-3}+\frac{1}{4^3}\left(\frac{2}{3}\right)^{n-3},$$
$$\cdot$$
$$\cdot$$
$$\cdot$$
$$\frac{1}{4^{n-2}}a_2=\frac{1}{4^{n-1}}a_1+\frac{1}{4^{n-1}}\left(\frac{2}{3}\right)^{1}.$$
Id est, $$a_n=\frac{1}{4^{n-1}}a_1+\frac{1}{4}\left(\frac{2}{3}\right)^{n-1}+...+\frac{1}{4^{n-1}}\left(\frac{2}{3}\right)^{1}=$$
$$=\frac{1}{4^{n-1}}a_1+\frac{\frac{1}{4}\left(\frac{2}{3}\right)^{n-1}\left(\left(\frac{3}{8}\right)^{n-1}-1\right)}{\frac{3}{8}-1}=\frac{a_1}{4^{n-1}}+\frac{2}{5}\left(\left(\frac{2}{3}\right)^{n-1}-\left(\frac{1}{4}\right)^{n-1}\right).$$
