Your equation is a second order linear equation so it should be a two-dimensional space of solutions (that is, solutions that depend on two free parameters) while your final answer depends only on one free parameter $C_0$ so this means you did something wrong.
Indeed, your equations don't determine what is $C_1$ which actually means that $C_1$ can be arbitrary. In addition, you made a mistake in deducing the general pattern. For example,
$$ C_6 = \frac{3}{6 \cdot 5} C_4 = -\frac{3}{6 \cdot 5} \frac{1}{4!} C_0 = -\frac{3C_0}{6!} \neq -\frac{C_0}{6!}.$$
In fact, we have
$$ C_{2m} = \frac{(2m-3)C_{2m-2}}{(2m)(2m-1)} = \frac{(2m-3)(2m-5)C_{2m-4}}{(2m)(2m-1)(2m-2)(2m-3)} = \dots = -\frac{(2m - 3)(2m - 5) \dots 1}{(2m)!} C_0 = - \frac{(2m - 3)(2m - 5) \dots 1}{(2m)(2m-1)(2m-2)(2m-3)(2m-4) \dots 1} C_0 = -\frac{C_0}{(2m-1)2^{m}m!}$$
and the general solution is given by
$$ y(x) = C_1 \cdot x - C_0 \sum_{m=0}^{\infty} \frac{x^{2m}}{(2m-1)2^m m!}. $$
Every time I see a question related to solving ODEs with varying coefficients via power series expansions I have a gut feeling it is cheating. The coefficients of the ODE are usually chosen in such a way that the power series ansatz leads to a simple recursion i.e. such that involves two terms only and can easily be solved in close form. This example is however different since as we can see the recursion involves three terms and we do not know how to solve it. Therefore we need to resort to a different method which I will outline right now.
As we know every linear second order ODE of the form:
\begin{equation}
y^{''}(x) + a_1(x) y^{'}(x) + a_2(x) y(x) = 0
\end{equation}
can be reduced to the one without a term involving the first derivative by the following assumption $y(x) := \exp(-1/2 \int a_1(x) dx) v(x)$ with the function $v$ satisfying the following ODE:
\begin{equation}
v^{''}(x) + Q(x) v(x) =0
\end{equation}
where $Q(x) := \left(-\frac{1}{4} a_1(x)^2 +a_2(x) -\frac{1}{2} a_1^{'}(x) \right)$.
In our case $a_1(x):= (x-2)/(x+1)$ and $a_2(x) := 1/(x+1)$ and $y(x):= \exp(-x/2) (1+x)^{3/2} v(x)$ and therefore $Q(x) = (-15+10 u-u^2)/(4 u^2)$ with $u:=(x+1)$.
This however resembles the confluent hypergeometric equation . Indeed an equation of the form:
\begin{equation}
v^{''}(x) + \left( \frac{-b^2+2 b+x(-4 a+2 b)-x^2}{4 x^2}\right)v(x)=0
\end{equation}
is solved by $x^{b/2} \exp(-x/2)( C_1 U(a,b,x)+ C_2 M(a,b,x) )$ .
We therefore solve the equations:
\begin{eqnarray}
-15 &=& -b^2+2 b\\
+10&=& -4 a+2 b
\end{eqnarray}
which gives us $(a,b)=(-4,-3)$ or $(a,b)=(0,5)$. The first solution is spurious and the second leads to a result.
We have:
\begin{equation}
y(x) = \exp(-x) \cdot (1+x)^4
\end{equation}
The second independent solution needs to be found still. It can be found using the Wronskian though.
Best Answer
$$\begin{equation}\sum_{n=2}^\infty n(n-1)c_nx^{n-2}-\sum_{n=1}^\infty nc_nx^{n-1}=0 \end{equation}$$ Change all the indices: $$\begin{equation}\sum_{n=0}^\infty (n+2)(n+1)c_{n+2}x^{n}-\sum_{n=0}^\infty (n+1)c_{n+1}x^{n}=0 \end{equation}$$ $$\begin{equation}\sum_{n=0}^\infty ((n+2)(n+1)c_{n+2}-(n+1)c_{n+1})x^{n}=0 \end{equation}$$ Hence : $$(n+2)c_{n+2}=c_{n+1} \text { for } n \ge 0$$ $$\implies c_n=\dfrac {c_1}{n!} \text { for } n \ge 1$$ The solution is : $$y(x)=\sum_{n=0}^\infty c_nx^n$$ $$y(x)=c_0+\sum_{n=1}^\infty c_nx^n=c_0+\sum_{n=1}^\infty \dfrac {c_1}{n!}x^n$$ $$y(x)=c_0-c_1+\sum_{\color{red}{n=0}}^\infty \dfrac {c_1}{n!}x^n =\underbrace {c_0-c_1}_{\text {=constant } C}+c_1e^x $$ $$\boxed {y(x)=C+C_1e^x}$$ The second solution is simply a constant.