How would you solve this recurrence equation: $a_{n+1}-2a_{n}=6\cdot 5^n$ for $n\geq 1$ How would you solve $a_{n+1}-2a_{n}=6\cdot 5^n$ for $n\geq 1$ ?
I don't understand the text in my textbook. I Would like somebody to explain it to me.
 A: Our recurrence is linear. The  theory of linear recurrences tells us that the general solution of our recurrence has shape $a_n=G(n)+P(n)$, where $G(n)$ is the general solution of the homogeneous recurrence $a_{n+1}-2a_n=0$, and $P(n)$ is some fixed particular solution of our given recurrence.
The homogeneos recurrence $a_{n+1}-2a_n$ is simple to handle. Rewrite it as $a_{n+1}=2a_n$. This says that $a_n$ doubles  we increment $n$ by $1$.
Thus $G(n)=(C)(2^n)$ for some constant $C$.
A particular solution $P(n)$ can be a bit harder to find. Let's look for a solution of the shape $P(n)=(k)(5^n)$. Substituting in our original equation, we get
$$(k)(5^{n+1})-(2k)(5^n)=(6)(5^n).\tag{$1$}$$
Note that $(k)(5^{n+1})=(5a)(5^n)$. Substitute in $(1)$, and cancel the $5^n$. We get $3k=6$ and therefore $k=2$. Thus $(2)(5^n)$ is a particular solution of our recurrence. 
So the general solution of our original recurrence is 
$$a_n=(C)(2^n)+(2)(5^n),$$
where $C$ is an arbitrary constant. If in addition we are told what $a_1$ is, we can find $C$. 
A: Using Generating Functions: Let $\displaystyle A(x)=\sum_{n\geq0}a_nx^n$. Then multiply your recurrence relation by $x^n$ and then take the sum it for $n\geq0$ to get
$$
\sum_{n\geq0}a_{n+1}x^n-\sum_{n\geq0}2a_nx^n=\sum_{n\geq0}6\cdot5^nx^n
$$
Now
$$
\begin{align}
\sum_{n\geq0}a_{n+1}x^n&=\frac{1}{x}\sum_{n\geq0}a_{n+1}x^{n+1}\\
&=\frac{1}{x}\sum_{n\geq1}a_{n}x^n\\
&=\frac{1}{x}\left(\sum_{n\geq0}a_{n}x^n-a_0\right)\\
&=\frac{1}{x}A(x)-\frac{a_0}{x}
\end{align}
$$
It is easy to see that
$$
\sum_{n\geq0}2a_nx^n=2A(x)
$$
and
$$
\sum_{n\geq0}6\cdot5^nx^n=6\sum_{n\geq0}(5x)^n=\frac{6}{1-5x}
$$
So you have
$$
\frac{1}{x}A(x)-\frac{a_0}{x}-2A(x)=\frac{6}{1-5x}
$$
If you solve this for $A(x)$, you get
$$
A(x)=\frac{6x}{(1-5x)(1-2x)}+\frac{a_0}{1-2x}
$$
Now observe that
$$
\begin{align}
\frac{6x}{(1-5x)(1-2x)}&=-\frac{2}{1-2x}+\frac{2}{1-5x}\\
&=-2\sum_{n\geq0}(2x)^n+2\sum_{\geq0}(5x)^n\\
&=\sum_{n\geq0}\left(-2^{n+1}+2\cdot5^n\right)x^n
\end{align}
$$
Also it is easy to see that
$$
\frac{a_0}{1-2x}=\sum_{n\geq0}a_02^nx^n
$$
Combining these, we have
$$
A(x)=\sum_{n\geq0}\left(-2^{n+1}+2\cdot5^n+a_02^n\right)x^n
$$
Therefore
$$
a_n=(a_0-2)2^n+2\cdot5^n
$$
A: \begin{align}
a_{n+1} & = 6 \cdot 5^n + 2 a_n\\
& = 6 \cdot 5^n + 2 (6 \cdot 5^{n-1} + 2a_{n-1})\\
& = 6 \cdot 5^n + 2 \cdot 6 \cdot 5^{n-1} + 2^2a_{n-1}\\
& = 6 \cdot 5^n + 2 \cdot 6 \cdot 5^{n-1} + 2^2 (6 \cdot 5^{n-2} + 2a_{n-2})\\
& = 6 \cdot 5^n + 2 \cdot 6 \cdot 5^{n-1} + 2^2 \cdot 6 \cdot 5^{n-2} + 2^3a_{n-2}\\
& = 6 \cdot 5^n + 2 \cdot 6 \cdot 5^{n-1} + 2^2 \cdot 6 \cdot 5^{n-2} + 2^3(6 \cdot 5^{n-3} + 2 a_{n-3})\\
& = 6 \cdot 5^n + 2 \cdot 6 \cdot 5^{n-1} + 2^2 \cdot 6 \cdot 5^{n-2} + 2^3 \cdot 6 \cdot 5^{n-3} + 2^4 \cdot a_{n-3}
\end{align}
Hence, in general (you need to prove the claim below using induction)
$$a_{n+1} = 2^{j+1}a_{n-j} + 6 \left( \sum_{k=0}^{j} 2^k 5^{n-k} \right)$$
Setting $j=n$, we get that
\begin{align}
a_{n+1} & = 2^{n+1}a_0 + 6 \left( \sum_{k=0}^{n} 2^k 5^{n-k} \right) = 2^{n+1}a_0 + 6 \cdot 5^n \left( \sum_{k=0}^{n} \left(\dfrac25 \right)^k \right)\\
& = 2^{n+1}a_0 + 6 \cdot 5^n \cdot \dfrac{1-(2/5)^{n+1}}{1-2/5}\\
& = 2^{n+1}a_0 + 6 \cdot \dfrac{5^{n+1} -2^{n+1}}3\\
& = 2^{n+1}a_0 + 2 \cdot (5^{n+1} -2^{n+1})
\end{align}
EDIT
You can verify if it is true by plugging the value for $a_n$ and $a_{n+1}$ and computing $a_{n+1} - 2a_n$.
$$a_n = 2^{n}a_0 + 2 \cdot (5^{n} -2^{n})$$
Hence, $$2a_n = 2^{n+1}a_0 + 4 \cdot (5^{n} -2^{n})$$
Hence,
\begin{align}
a_{n+1} - 2a_n & = 2^{n+1}a_0 + 2 \cdot (5^{n+1} -2^{n+1}) - \left( 2^{n+1}a_0 + 4 \cdot (5^{n} -2^{n})\right)\\
& = 2 \cdot (5^{n+1} -2^{n+1}) - 4 \cdot (5^n - 2^n)\\
& = 2 \cdot 5^{n+1} -2^{n+2} - 4 \cdot 5^n + 4 \cdot 2^n\\
& = 10 \cdot 5^{n} - 4 \cdot 5^n\\
& = 6 \cdot 5^n
\end{align}
A: A Functional Calculus Approach
Let $S$ be the shift operator; i.e. $S(f(n))=f(n+1)$.
$$
(S-2)a(n)=6\cdot5^n
$$
Formally, the inverse of $S-2$ is
$$
-\frac12(1+S/2+S^2/4+S^3/8+\dots)
$$
Unfortunately, this does not converge for the given sequence, so let's consider $1-2S^{-1}$:
$$
(1-2S^{-1})a(n)=6\cdot5^{n-1}
$$
Formally, the inverse of $1-2S^{-1}$ is
$$
1+2S^{-1}+4S^{-2}+8S^{-3}+\dots
$$
Applying this gives a particular solution
$$
\begin{align}
a(n)
&=6\cdot5^{n-1}+2\cdot6\cdot5^{n-2}+4\cdot6\cdot5^{n-3}+8\cdot6\cdot5^{n-3}+\dots\\
&=6\cdot5^{n-1}\left(1+\frac25+\frac4{25}+\frac8{125}+\dots\right)\\
&=6\cdot5^{n-1}\frac1{1-2/5}\\
&=2\cdot5^n
\end{align}
$$
We also have the homogeneous solution of $C\,2^n$ for $(S-2)a(n)=0$, so the general solution is
$$
a_n=2\cdot5^n+C\,2^n
$$

A Geometric Series Approach
Let $a_n=2^n\,b_n$. Then the recurrence becomes
$$
2^{n+1}b_{n+1}-2^{n+1}b_n=6\cdot5^{n}
$$
That is
$$
b_{n+1}-b_n=3\cdot(5/2)^n
$$
Using the formula for the sum of a geometric series, we get that
$$
b_{n+1}-b_0=\frac{3\cdot(5/2)^{n+1}-3}{5/2-1}
$$
which gives
$$
b_n=2\cdot(5/2)^n+C
$$
and reverting to $a_n$:
$$
a_n=2\cdot5^n+C\,2^n
$$
