Pattern Recognition - How to solve this problem? A general term of the sequence
$$
-8,-7,-10,-1,-28,53,\ldots
$$
can be expressed as $\dfrac{a-b^{n-1}}{4}$, where $a$ and $b$ are integers.
What is the value of $ab$?
Teach me how to solve this problem.
 A: $\frac{a-b^{n-1}}{4}=-8 (\text{for } n=1)$, or $\frac{a-1}{4}=-8 \implies a=-31$, as $n=1$ gives us first term of the series,
similarly for $n=2$ the expression must be equal to $-7$, or $\frac{-31-b}{4}=-7 \implies b=-3$
A: For $n=1$:
$$\dfrac {a-1}4=-8$$
and for $n=2$:
$$\frac{a-b}4=-7$$
Solve the system, check that the formula works for the other terms and you have $a$ and $b$.
A: If we call this sequence $x_n$, then $a-4x_n$ is the geometric sequence $b^{n-1}$. So the first element should be $1$, meaning $a-4(-8)=1$, and the second element should be $b$, so $b=a-4(-7)$.
A: Suppose you did not know that the sequence started at $n=1$ and instead just had three consecutive terms $\ldots ,x_n,x_{n+1},x_{n+2}, \ldots$ with $n$ as yet unknown:  
You could use $b=\dfrac{\dfrac{a-b^{n+1}}{4}-\dfrac{a-b^{n}}{4}}{\dfrac{a-b^{n}}{4}-\dfrac{a-b^{n-1}}{4}}$ and then $a= \dfrac4{b-1}\left(b\dfrac{a-b^{n-1}}{4} - \dfrac{a-b^{n}}{4}   \right)$ 
i.e. $b=\dfrac{x_{n+2}-x_{n+1}}{x_{n+1}-x_{n}}$ and then $a= \dfrac4{b-1}\left(bx_{n} - x_{n+1}   \right)$. 
