Find $a(n)$ with generating functions of recurrence: \begin{cases} a(1)=3 \;,\;a(2)=10 \\ a(n)=6a(n-1)-3a(n-2)\end{cases} I went till : $h(x)-10x-3=6xh(x)-3x^2h(x)$
but I'm stuck in the part of partial fractions. Please show your work till the end of solution.
Assuming
$$h(x)=\sum_{n\ge1}a(n)x^{n-1}=a(1)+a(2)x+a(3)x^2+\cdots$$
is the generating function for $a(n)$, we have
$$\begin{align*} a(n)&=6a(n-1)-3a(n-2)\\[1ex] \sum_{n\ge3}a(n)x^{n-1}&=6\sum_{n\ge3}a(n-1)x^{n-1}-3\sum_{n\ge3}a(n-2)x^{n-1}\\[1ex] &=6x\sum_{n\ge3}a(n-1)x^{n-2}-3x^2\sum_{n\ge3}a(n-2)x^{n-3}\\[1ex] &=6x\sum_{n\ge2}a(n)x^{n-1}-3x^2\sum_{n\ge1}a(n)x^{n-1}\\[1ex] h(x)-a(2)x-a(1)&=6x\big(h(x)-a(1)\big)-3x^2h(x)\\[1ex] (1-6x+3x^2)h(x)&=3-8x\\[1ex] h(x)&=\frac{3-8x}{1-6x+3x^2} \end{align*}$$
For the partial fraction decomposition, you can use the quadratic formula to show that the denominator has roots at $x=1\pm\sqrt{\frac23}$.
Then
$$\begin{align*} h(x)&=\frac{\theta_1}{x-\left(1-\sqrt{\frac23}\right)}+\frac{\theta_2}{x-\left(1+\sqrt{\frac23}\right)}\\[1ex] 3-8x&=\theta_1\left(x-\left(1+\sqrt{\frac23}\right)\right)+\theta_2\left(x-\left(1-\sqrt{\frac23}\right)\right) \end{align*}$$
Using the "cover-up" method, we have
$$x=1-\sqrt{\frac23}\implies\theta_1=\frac52\sqrt{\frac32}-4$$
$$x=1+\sqrt{\frac23}\implies\theta_2=-\frac52\sqrt{\frac32}-4$$