Find closed form for $a_n=2\frac{n-1}{n}a_{n-1}-2\frac{n-2}{n}a_{n-2}$ for all $n \ge 3$ Find a closed form for:
$a_n=2\frac{n-1}{n}a_{n-1}-2\frac{n-2}{n}a_{n-2}$ for all $n \ge 3$
Initial values: $a_1=3, a_2=1$
I'm struggling a bit with the algebra as the solution contains complex number. Any help would be appreciated.
 A: Remark that it satisfies
$$
na_n=2\left(\left(n-1\right)a_{n-1}-\left(n-2\right)a_{n-2}\right)
$$
Then I recommend you study the sequence $\left(b_n\right)_{n \in \mathbb{N}}$ defined by
$$
b_n=na_n
$$
which satisfies
$$
b_n=2\left(b_{n-1}-b_{n-2}\right)
$$
Do you know how to deal with this kind of sequence ?
EDIT : 
The sequence is what we call linear induction sequence of order $2$ that I prefer to rewrite for $n \in \mathbb{N}^{*}$
$$
b_{n+2}-2b_{n+1}+2b_n=0
$$
You search for geometrical sequences that satisfy it, it leads you to solve the equation
$$
r^2-2r+2=0
$$
The discriminant equals to $\Delta=4-8=-4<0$. Here comes the complex solution that you were talking about ;  the two solutions are 
$$
r_1=\frac{2+2i}{2}=1+i \  \text{ and } \ r_2=\frac{2-2i}{2}=1-i
$$
Then those two geometrics are solutions, and it is a two-dimensional space so we got the solution
$$
b_n=\alpha\left(1+i\right)^{n-1}+\beta \left(1-i\right)^{n-1}
$$
So using $b_n=na_n$ it comes that your sequence that looks not easy to compute is in fact verifying for all $n \in \mathbb{N}$
$$
a_n=\frac{\alpha\left(1+i\right)^{n-1}+\beta \left(1-i\right)^{n-1}}{n}
$$
Then $a_1=3$ and $a_2=1$ so
$$
\alpha+\beta=3 \ \text{ and } \ 2=\alpha\left(1+i\right)+\beta\left(1-i\right)
$$
You inject the first in the second
$$
2=\alpha\left(1+i\right)+\left(3-\alpha\right)\left(1-i\right) \Leftrightarrow 2i\alpha=3i-1 \Leftrightarrow \alpha=\frac{3}{2}+\frac{i}{2}
$$
And $$\beta=3-\alpha=\frac{3}{2}-\frac{i}{2}$$
Hence this dreadful sequence is given for $n \in \mathbb{N}^{*}$ by

$$
a_n=\left(\frac{3}{2}+\frac{i}{2}\right)\frac{\left(1+i\right)^{n-1}}{n}+\left(\frac{3}{2}-\frac{i}{2}\right)\frac{\left(1-i\right)^{n-1}}{n}
$$

Hope it helped you, if you have any question on how to proceed, dont hesitate.
A: Set $u_n = na_n $ then your relation becomes 
$$u_n = 2u_{n-1}-2u_{n-2}$$
Whose characteritic equation is $$x^2-2x+2=0 \implies x=\frac{2\pm 2i}{2}$$
And hence $$ u_n =A\left(\frac{2-2i}{2}\right)^{n-1} +B\left(\frac{2+2i}{2}\right)^{n-1}$$

that is $$\color{red}{a_n =\frac{A}{n}\left(\frac{2-2i}{2}\right)^{n-1} +\frac{B}{n}\left(\frac{2+2i}{2}\right)^{n-1}}$$
  Now can you fine A and B?

A: My approach is similar to the others but somewhat easier to facilitate. We start off, of course, by taking $f_n=na_n$ so that we have the sequence
$$f_n=2f_{n-1}-2f_{n-2}$$
This is a generalized Fibonacci form, i.e., $f_n=a f_{n-1}+bf_{n-2}$, for which I have previously shown here that a general solution is given by
$$f_n=\left(f_1-\frac{af_0}{2}\right) \frac{\alpha^n-\beta^n}{\alpha-\beta}+\frac{af_0}{2} \frac{\alpha^n+\beta^n}{\alpha+\beta}=\frac{(f_1-f_0\beta)\alpha^n-(f_1-f_0\alpha)\beta^n}{\alpha-\beta}\quad n=0,1,2,\cdots$$
or
$$f_n=\frac{(f_2-f_2\beta)\alpha^{n-1}-(f_2-f_1\alpha)\beta^{n-1}}{\alpha-\beta}\quad n=1,2,3,\cdots$$
where $\alpha,\beta$ are the characteristic roots given by
$$\alpha,\beta=(a\pm\sqrt{a^2+4b})/2$$
Thus, substituting $f_1=a_1$ and $f_2=2a_2$, along with $\alpha,\beta=1\pm i$ into the above, and then $a_n=f_n/n$, I find that
$$a_n=\frac{(-1+3i)(1+i)^{n-1}-(-1-3i)(1-i)^{n-1}}{2ni}$$
I have verified this solution numerically.
