How to find $n$-th value in a series Let $(x_n, y_n, z_n) = (3, 1, 0)$ for $n=0$
For $n \ge 1$,
$$\begin{align}
x_n &= x_{n-1} +3 z_{n-1}\\
y_n &= x_{n-1} +2 z_{n-1}\\
z_n &= 5 y_{n-1}
\end{align}$$
Please let me know the formula to find $x_n,y_n,z_n$ values of any integer $n$.
The following would be the series:
$$\begin{array}{c:l}
n & x_n,y_n,z_n\\
\hline
0 & 3,1,0\\
1 & 3,3,5\\
2 & 18,13,15\\
3 & 63,48,65\\
\end{array}$$

As pointed out by GerryMyerson, you can rewrite your problem as
\begin{align}
X_n=A\cdot X_{n-1} =\ldots=A^nX_0
\end{align}
With $X_n =(x_n,y_n,z_n)^T$ and $X_0=(3,1,0)$, we only need the appropriate matrix $A$. 
Your recurrence relation is 
\begin{align}
x_n &=x_{n-1}+3z_{n-1}\\
y_n &=x_{n-1}+2z_{n-1}\\
z_n &= 5 y_{n-1}
\end{align}
Or as matrix-vector product
\begin{align}
\begin{pmatrix}x_{n-1} \\ y_{n-1}\\ z_{n-1} \end{pmatrix}
=
\underbrace{\begin{pmatrix}1 & 0 & 3\\ 1 &  0 & 2 \\ 0 & 5 &0 \end{pmatrix}}_{=:A}
\begin{pmatrix}x_{n-1} \\y_{n-1}\\z_{n-1} \end{pmatrix}
\end{align}
Now we have to diagonalize $A$, i.e. finding matrices, such that $A=V D V^T$ 
Do you know how to proceed?
