How does recurrence relation works with system of equations? There are      $$   a+b+c+d = 2\\2a+2^2b+2^3c+2^4d = 5\\ 3a+3^2b+3^3c+3^4d = 6\\4a+4^2b+4^3c+4^4d = 1$$
then I'm given 
$$C_{n}= a+bn+cn^2+dn^3$$
from linear recurrence relation with repeated roots said that

$$(x-1)^4 = \sum_{k=0}^{4}\binom{4}{k}(-1)^{4-k}x^{4-k}$$

and then   
$$C_{n}=\sum_{k=1}^{4}\binom{4}{k}(-1)^{5-k}C_{n-k} = 4C_{n-1}-6C_{n-2}+4C_{n-3}-C_{n-4}$$ $\forall n \geq 5$
I want to find $C_{6}$. Then i've plugged $C_{5}=-3$ into that relation which finally gave $C_{6}=-8$ but wolfram alpha told me that $C_{6} =-48 
Question: How do I use recurrence relations to find out the correct $C_6$
 A: I believe you are over-complicating it. The given system tells you that the quartic polynomial $p(n)=an+bn^2+cn^3+dn^4$ attains the values $2,5,6,1$ at $n=1,2,3,4$. By Lagrange interpolation such polynomial is 
$$ p(n) = \frac{1}{24}\left(18n+37n^2-6n^3-n^4\right) $$
hence
$$ C_6 = \frac{p(6)}{6} = \color{red}{-8}.$$
A: 
How do I use recurrence relations to find out the correct $C_6$

You did find the correct $\,C_6 = -8\,$.

$\;C_{n}== 4C_{n-1}-6C_{n-2}+4C_{n-3}-C_{n-4} \quad\quad\forall n \geq 5$

Correct, thus far.

but wolfram alpha told me that $C_{6} =-48$

What did you ask WA?
WA's answer to $\,c_{n} = 4 c_{n-1} - 6 c_{n-2} + 4 c_{n-3} - c_{n-4}\,$$\;\style{font-family:inherit}{\text{with}}\;$$\,c_1 = 2\,,$$\,c_2 = 5/2\,,$$\,c_3 = 2\,,$$\,c_4 = 1/4\,$:
$$
C_n = \frac{3}{4} - \frac{1}{24} n \big(n \left(n + 6\right) - 37\big) = \frac{-n^3 - 6 n^2 + 37 n + 18}{24}
$$
Unsurprisingly, this matches $\,\frac{p(n)}{n}\,$ from Jack D'Aurizio's answer, and $\,C_5=-3, C_6=-8\,$ indeed.
However, if you are maybe looking for $\,6a+6^2b+6^3c+6^4d\,$, instead, then that's $\;6 \cdot C_6 = -48\,$.
