\section{(Multiple roots of Characteristic Polynomial)}
The characteristic polynomial is
\[\rho(\xi) = \sum_{j=0}^{k}\alpha_{j}\xi^{j}\]
with the corresponding difference equation
\begin{eqnarray}
\sum_{j=0}^{k}\alpha_{j}y_{n+j} & = & 0
\end{eqnarray}
$\therefore$
\[\rho(r) = \rho '(r) = \cdots = \rho^{(q-1)}(r) = 0\]
Now
\[\rho(r) = 0 \Rightarrow \sum_{j=0}^{k}\alpha_{j}r^{j} = 0\]
Comparing with (1) we say that $y_{n} = r^{n}$ is the solution of the difference equation.
\begin{eqnarray*}
 && \rho'(r) = 0 \\
\Rightarrow && \sum_{j=0}^{k}j\alpha_{j}r^{j-1} = 0 \\
\therefore && \frac{\sum_{j=0}^{k}j\alpha_{j}r^{j}}{r} = 0 \\
\therefore && \sum_{j=0}^{k}j\alpha_{j}r^{j} = 0
\end{eqnarray*}
Again, comparing with eqn (1), we get that $y_{n} = nr^{n}$ is a solution of the difference equation.
\begin{eqnarray*}
&&\rho''(r) = 0 \\
\Rightarrow && \sum_{j=0}^{k}j^{2}\alpha_{j}r^{j-2} = 0 \\
\therefore && \frac{\sum_{j=0}^{k}j^{2}\alpha_{j}r^{j}}{r^{2}} = 0 \\
\therefore && \sum_{j=0}^{k}j^{2}\alpha_{j}r^{j} = 0
\end{eqnarray*}
Again comparing with (1) we get that $y_{n} = n^{2}r^{n}$ is the solution.  Similarly
\begin{eqnarray*}
&& \rho^{(q-1)}(r) = 0 \\
\Rightarrow && \sum_{j=0}^{k}j^{q-1}\alpha_{j}r^{j-[q-1]} = 0 \\
\therefore && \frac{\sum_{j=0}^{k}j^{q-1}\alpha_{j}r^{j}}{r^{q-1}} = 0 \\
\therefore && \sum_{j=0}^{k}j^{q-1}\alpha_{j}r^{j} = 0
\end{eqnarray*}
Comparing with (1) we get $y_{n} = n^{q-1}r^{n}$ is the solution of the difference equation.