Find the general form of sequence $(a_n)$ satisfying the following equation: Find the general form of sequence $(a_n)$ satisfying the following equation:
$3a_{n+1} − 5a_n = 5^n − 3^n + 2^{n+1} − 4$
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What's the general way to solve exercises like that?
 A: I would solve it with generating functions, personally. I'll add a short description, but can elaborate if you'd like. 
Briefly, if we set $A(x) = \sum_n a_n x^n$, then we can multiply both sides of your recurrance by every $x^n$, and sum these up. This gives us
$$3 \sum_n a_{n+1} x^n - 5 \sum_n a_n x^n = \sum_n 5^n x^n - \sum_n 3^n x^n + \sum_n 2^{n+1} x_n - 4 \sum x^n$$
After recognizing series on the right hand side as geometric, and substituting our generating function $A$ on the left, we see
$$3 \frac{A-a_0}{x} - 5A = \frac{1}{1-5x} - \frac{1}{1-3x} + 2 \frac{1}{1-2x} - 4 \frac{1}{1-x}$$
It is now a matter of algebra to solve for $A(x)$, and then perform a partial fraction decomposition or similar to find a closed form. 
Edit:
Let's go into more detail both in how we got here, and where we go. 
As for how we got here, I think the right side is self explanatory. The left side, however, requires a trick. I think it is best seen right to left:
$$
\frac{A - a_0}{x} =
\frac{ \sum_n a_n x^n - a_0}{x} =
\frac{\sum_{n \geq 1} a_n x^n}{x} =
\sum_{n \geq 1} a_n x^{n-1} =
\sum_m a_{m+1} x^m
$$
The main idea is to cancel the $a_0$ term, so that we can factor out an $x$. Doing this move shifts the entire sequence left, and then we reindex so it is easier to work with. 
Now that we better understand how we got here, let's shift focus to finishing strong. We can rearrange the above formula to get
$$
A = 
\frac{3a_0}{3-5x} + 
\left (
\frac{1}{3-5x} 
\right )
\left (
\sum_n (5^n - 3^n + 2^{n+1} - 4) x^{n+1}
\right )
$$
Here we've re-expanded the geometric series on the right side, and we have an $x^{n+1}$ because we multiplied both sides by $x$.
If we write $\frac{1}{3-5x}$ as 
$\frac{1}{3} \frac{1}{1-\frac{5}{3}x}$, we recognize a second geometric series, which we now expand:
$$
A =
a_0 \sum_n \left ( \frac{5}{3} \right )^n x^n
+
\frac{1}{3}
\left (
\sum_n \left ( \frac{5}{3} \right )^n x^n
\right )
\left (
\sum_n (5^n - 3^n + 2^{n+1} - 4) x^{n+1}
\right )
$$
Now we see the light! If we can find the $x^n$ term in the power series on the right, we're done! But that coefficient is just $a_0 \frac{5}{3}^n$ plus the coefficient of the product. 
Thankfully, we remember (or at least Wikipedia does) how to compute the product of two power series. If you don't remember, foil the first few terms and check if you can see the pattern (then prove it :P ). At the end of the day, we get the following closed form:
$$a_n = 
a_0 \left ( \frac{5}{3} \right )^n 
+
\frac{1}{3} 
\sum_{i+j=n, j \neq 0} \left ( \frac{5}{3} \right )^i (5^{j-1} - 3^{j-1} + 2^j - 4)
$$
Where the $j \neq 0$ comes from that series having $0$ as its $x^0$ coefficient. We don't want a $(5^{-1} - 3^{-1} + 2^0 - 4)$ term. 

Now, granted, this closed form is gross. But so was the original recurrance, so ¯_(ツ)_/¯. 
If we use $a_0 = 0$ and compute the first few terms, we'll see that the recurrence and our closed forms agree:
$a_0 = 0; a_1 = 0; a_2 = \frac{2}{3}$, and so on. 
If you want practice working with these kinds of problems, or you want too see a better exposition (I don't have space to write as much detail as I would like here), you should read Generatingfunctionology, which is freely available online. I mentioned this book in the comments, but it's such a great read I felt obligated to mention it here too. 

I hope this helps ^_^
