Finding the general term for the sequence $a_n = \frac{3}{4}a_{n-1} +4e$ How do I find the general term for the sequence $$a_n = \frac{3}{4}a_{n-1} +4e$$ using a generating function? If there is an easier way to do it without using a generating function, please tell me. Thanks in advance.
 A: As Did notes in comments above, getting a generating function is a bit of overkill here. That said, I figured I'd show how it works in this case for both exponential and ordinary generating functions.

Exponential GF
We introduce $A_e (x)=\sum\limits_{n=0}^\infty \dfrac{a_n}{n!}x^n$ as the exponential generating function of $\{a_n\}$. Then
\begin{align}
A_e'(x) 
 = \sum_{n=1}^\infty \frac{a_n}{(n-1)!}x^{n-1}
&= \frac34 \sum_{n=1}^\infty \frac{a_{n-1}}{(n-1)!}x^{n-1}
+e\sum_{n=1}^\infty \frac{1}{(n-1)!}x^{n-1}\\
&= \frac34 \sum_{n=0}^\infty \frac{a_{n}}{(n)!}x^{n}
+e\sum_{n=0}^\infty \frac{1}{(n)!}x^{n}\\
&=\frac34 A_e(x)+e^{x+1}
\end{align}
So now we have an inhomogeneous ODE for $A_e(x)$. If we solve this, then we can identify $a_n=A_e^{(n)}(0)$ and obtain the sequence.

Ordinary GF
We introduce $A_o(x)=\sum\limits_{n=0}^\infty a_n x^n$ as the ordinary generating function of $\{a_n\}$. Then
\begin{align}
A_o(x)
&=a_0+\sum_{n=1}^\infty a_n x^n\\
&=a_0+\frac34\sum_{n=1}^\infty a_{n-1} x^n+e\sum_{n=1}^\infty x^n\\
&=a_0+\frac34 x A_o(x)+\frac{ex}{1-x}
\end{align}
We can solve this equation for $A_o(x)$ and expand in a power series to obtain $a_n$.
A: Generating functions are a good method, but for this straightforward problem there's another one, since you can get rid of the constant:
$$
a_ n = \frac{3}{4} a_{n-1} + 4e\\
a_{n-1} = \frac{3}{4}a_{n-2}  +4e\\
a_n - a_{n-1} = \frac{3}{4}(a_{n-1} -a_{n-2})
$$
Now set $\Delta a_n  = a_n - a_{n-1}$
$$
\Delta a_n = \frac{3}{4} \Delta a_{n-1} = \bigg(\frac{3}{4}\bigg)^2 \Delta a_{n-2} = \ldots =\bigg(\frac{3}{4}\bigg)^{n-1} \Delta a_1
$$
This get s you a telescoping sum on the LHS and a geometric sum over $n$ on the RHS. 
A: I'd write it as
$$
a_{n+1}=\frac{3}{4}a_n+4e
$$
so we can start counting from $0$. We also have
$$
a_{n+2}=\frac{3}{4}a_{n+1}+4e
$$
so
$$
a_{n+2}=\frac{3}{4}a_{n+1}+a_{n+1}-\frac{3}{4}a_n
$$
or
$$
4a_{n+2}-7a_{n+1}+3a_n=0
$$
which has, as characteristic polynomial $4X^2-7X+3$. The roots are $1$ and $3/4$, so the general solution is
$$
a_{n}=\alpha 1^n+\beta\left(\frac{3}{4}\right)^n
$$
with $a_1=3(a_0+16e)/4$ and $a_0$ arbitrary. Thus we get
$$
\begin{cases}
a_0=\alpha+\beta\\[2ex]
\dfrac{3}{4}a_0+4e=\alpha+\dfrac{3}{4}\beta
\end{cases}
$$
