Problem source

\begin{questionparts}
\item
The function $\f(x)$ is defined for $\vert x \vert < \frac15$ by 
\[
\f(x) = \sum_{n=0}^\infty a_n x^n\;,
\]
where $a_0=2$, $a_1=7$ and  $a_n =7a_{n-1} - 10a_{n-2}$ for $n\ge{2}\,$. 
Simplify $\f(x) - 7x\f(x) + 10x^2\f(x)\,$, and hence show that $\displaystyle\f(x) = {1\over 1-2x} + {1 \over 1-5x} \;$.
Hence show that $a_n=2^n + 5^n\,$.
\item The function $\g(x)$ is defined for $\vert x \vert < \frac13$ by 
\[
\g(x) = \sum_{n=0}^\infty b_n x^n \;,
\]
where $b_0=5\,$, $b_1 =10 \,$, $b_2=40\,$, $b_3=100$ and  $b_n = pb_{n-1} + qb_{n-2}$ for $n\ge{2}\,$. 
Obtain an expression for $\g(x)$ as the sum of two algebraic fractions and determine $b_n$ in terms of $n$.
\end{questionparts}

Solution source

\begin{questionparts}
\item
\begin{align*}
&& f(x) -7xf(x)+10x^2f(x) &= \sum_{n=0}^\infty a_n x^n - 7x \sum_{n=0}^{\infty} a_n x^n + 10x^2 \sum_{n=0}^{\infty} a_nx^n \\
&&&= \sum_{n=2}^\infty (a_n-7a_{n-1}+10a_{n-2})x^n + a_0+a_1x-7a_0x \\
&&&= 0 + 2-7x \\
\\
\Rightarrow && f(x) &= \frac{2-7x}{1-7x+10x^2} \\
&&&= \frac{2-7x}{(1-5x)(1-2x)} \\
&&&= \frac{1}{1-2x} + \frac{1}{1-5x} \\
&&&= \sum_{n=0}^{\infty} (2^n + 5^n)x^n
\end{align*}
Therefore $a_n = 2^n +5^n$

\item $\,$
\begin{align*}
&& 40 &= 10p + 5 q \\
&& 100 &= 40p+10q \\
&& 10 &= 4p + q \\
\Rightarrow && (p,q) &= (1,6) \\
\\
&& g(x) -xg(x)-6x^2g(x) &= 5+5x \\
\Rightarrow && g(x) &= \frac{5+5x}{1-x-6x^2} \\
&&&= \frac{5+5x}{(1-3x)(1+2x)} \\
&&&= \frac{4}{1-3x} + \frac{1}{1+2x} \\
&&&= \sum_{n=0}^{\infty} (4 \cdot 3^n + (-2)^n)x^n \\
\Rightarrow && b_n &= 4 \cdot 3^n + (-2)^n
\end{align*}
\end{questionparts}