How do I find the term of a recursive sequence? I have $\{a_n\}$ the following sequence:
$a_1 =-1$
$a_k a_{k+1} = - a_k - \frac{1}{4}$
How can I find $a_1 a_2 \cdots a_n$?
 A: $a_{k+1} = -\frac{4a_k+1}{4a_{k}}$
$a_2 = -3/4$
$a_3 = -2/3=-4/6$
$a_4 = -5/8$
$a_5 = -3/5 = -6/10$
$a_6 = -7/12$
So, we prove that $a_n = -(n+1)/(2n)$
Base case: $a_1 = -1$
IH: $a_{k} = -(k+1)/(2k)$
Then, $a_{k+1} = -\frac{4a_k+1}{4a_{k}} = -\frac{-2(k+1)/k + 1}{-2(k+1)/k} = -\frac{-2k-2+k}{-2(k+1)} = -((k+1)+1)/(2(k+1))$
Take it from here.
A: $$a_{k+1} = \frac{-4a_k - 1}{4a_k + 0}$$
Let $p_k / q_k = a_k$ :
$$\frac{p_{k+1}}{q_{k+1}} = \frac{-4 \frac{p_k}{q_k} - 1}{4\frac{p_k}{q_k} + 0} = \frac{-4 p_k - 1q_k}{4p_k + 0q_k}$$
$$\begin{align}
%
\begin{bmatrix} p_{k+1} \\ q_{k+1} \end{bmatrix} 
%
& = \begin{bmatrix} -4 & -1 \\ 4 & 0 \end{bmatrix} \begin{bmatrix} p_{k} \\ q_{k} \end{bmatrix} 
\\ %
& = \begin{bmatrix} -4 & -1 \\ 4 & 0 \end{bmatrix}^k \begin{bmatrix} p_{1} \\ q_{1} \end{bmatrix}
\\ %
& \text{ Jordan Decomposition...}
\\ %
& = \left(
\begin{bmatrix} -2 & 1 \\ 4 & 0 \end{bmatrix}
\begin{bmatrix} -2 & 1 \\ 0 & -2 \end{bmatrix}
\begin{bmatrix} -2 & 1 \\ 4 & 0 \end{bmatrix}^{-1}
\right)^k
\begin{bmatrix} -1 \\ 1 \end{bmatrix}
\\ % 
& =
\begin{bmatrix} -2 & 1 \\ 4 & 0 \end{bmatrix}
\begin{bmatrix} -2 & 1 \\ 0 & -2 \end{bmatrix}^k
\begin{bmatrix} -2 & 1 \\ 4 & 0 \end{bmatrix}^{-1}
\begin{bmatrix} -1 \\ 1 \end{bmatrix}
\\ % 
& =
\begin{bmatrix} -2 & 1 \\ 4 & 0 \end{bmatrix}
\begin{bmatrix} (-2)^k & k~(-2)^{k - 1} \\ 0 & (-2)^k \end{bmatrix}
\begin{bmatrix} -2 & 1 \\ 4 & 0 \end{bmatrix}^{-1}
\begin{bmatrix} -1 \\ 1 \end{bmatrix}
\\ % 
& =
(-2)^k
\begin{bmatrix} -k/2 - 1 \\ k + 1\end{bmatrix}
\\ % 
\end{align}$$
So 
$$a_{k+1} = \frac{p_{k+1}}{q_{k+1}} = \frac{-k/2 - 1}{k + 1}$$
$$a_k = \frac{-k - 1}{2k}$$
A: You can guess the answer calculating the first numbers:
$a_1=-1, a_2 = 1 -  \frac{1}{4a_1} = - 3/4, a_3 = - 2/3=-4/6, a_4=-5/8,...$
From this assume that $a_n = - \frac{(n+1)}{2n}$. You see immediately that $a_n \neq 0$ and that it fullfills the recursion.  
