Problem source

The real numbers $u_{0},u_{1},u_{2},\ldots$ satisfy the difference
equation 
\[
au_{n+2}+bu_{n+1}+cu_{n}=0\qquad(n=0,1,2,\ldots),
\]
where $a,b$ and $c$ are real numbers such that the quadratic equation
\[
ax^{2}+bx+c=0
\]
has two distinct real roots $\alpha$ and $\beta.$ Show that the above difference equation is satisfied by the numbers $u_{n}$ defined
by 
\[
u_{n}=A\alpha^{n}+B\beta^{n},
\]
where 
\[
A=\frac{u_{1}-\beta u_{0}}{\alpha-\beta}\qquad\mbox{ and }\qquad B=\frac{u_{1}-\alpha u_{0}}{\beta-\alpha}.
\]
Show also, by induction, that these numbers provide the only solution. 
Find the numbers $v_{n}$ $(n=0,1,2,\ldots)$ which satisfy 
\[
8(n+2)(n+1)v_{n+2}-2(n+3)(n+1)v_{n+1}-(n+3)(n+2)v_{n}=0
\]
with $v_{0}=0$ and $v_{1}=1.$

Solution source

First notice that $u_n = \alpha^n$ and $u_n = \beta^n$ both satisfy the recurrence, since:

\begin{align*}
&& a \alpha^2 + b \alpha + c &= 0 \\
\Rightarrow && a \alpha^{n+2} + b \alpha^{n+1} + c \alpha^n &= 0 \\
\Rightarrow && a u_{n+2} + bu_{n+1} + cu_n &=0
\end{align*}

Notice also that if $u_n$ and $v_n$ both satisfy the recurrence, then any linear combination of them will satisfy the recurrence:

\begin{align*}
&& \begin{cases} au_{n+2} + bu_{n+1} + cu_n &= 0 \\
av_{n+2} + bv_{n+1} + cv_n &= 0 \\ \end{cases} \\
\Rightarrow && a (\lambda u_{n+2}+ \mu v_{n+2}) +  b (\lambda u_{n+1}+ \mu v_{n+1}) +  c (\lambda u_{n}+ \mu v_{n}) &= 0
\end{align*}
by adding a linear combination of the top two equations.

Therefore it suffices to check that the constants $A$ and $B$ are such that we match $u_0$ and $u_1$.

$\frac{u_1 - \beta u_0}{\alpha - \beta} + \frac{u_1 - \alpha u_0}{\beta - \alpha} = u_0$ and $\frac{u_1 - \beta u_0}{\alpha - \beta}\alpha + \frac{u_1 - \alpha u_0}{\beta - \alpha}\beta = u_1$. So we are done.

Suppose we have another sequence, then we first notice that the first and second terms must be identical to each other. Suppose the first $k$ terms are identical, then since the $k+1$th term depends only on the $k$ and $k-1$th terms (both of which are equal) the $k+1$th term is the same. Therefore, by the principle of mathematical induction, all terms are the same. 

First notice that if you put $v_n = (n+1)w_n$ we have

\begin{align*}
&& 8(n+3)(n+2)(n+1)w_{n+2} - 2(n+3)(n+2)(n+1)w_{n+1} - (n+3)(n+2)(n+1)w_n &= 0 \\
\Rightarrow && 8w_{n+2}-2w_{n+1}-w_n &= 0
\end{align*}

This has characteristic equation $8\lambda^2 - 2\lambda - 1 = 0 \Rightarrow \lambda = \frac12, -\frac14$.

Therefore the general solution is $w_n = A \l \frac12 \r^n + B \l -\frac14\r^n$ and $v_n = (n+1)\l  A \l \frac12 \r^n + B \l -\frac14\r^n \r$.

When $n = 0$ we have $A+B = 0 \Rightarrow B =-A$.

When $n=1$ we have $1 = 2 \l \frac{A}{2} + \frac{A}{4} \r \Rightarrow A = \frac{4}{3}$, therefore

\[ v_n = \frac{4}{3}(n+1) \l \frac{1}{2^n} + \l -\frac14\r^n \r\]