Solve the recursion $a_n=\frac{1}{4}2^{n-1}-1+3a_{n-1}$ Solve the recursion $a_n=\frac{1}{4}2^{n-1}-1+3a_{n-1}$
I'd know how to solve it if it weren't for that -1. Because of it, I can't divide the particular equation with $2^{n-2}$ to solve it. What can be done here?
Oh, sorry. The starting conditions are $a_3=1$, and by "solve" I mean, find the explicit for of $a_n$ (the non recursive one)
 A: Hint:
Consider $b_n=a_n-\frac{1}{2}$
Can you write a recursive equation for $b_n$?
A: $\newcommand{\+}{^{\dagger}}%
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$a_{n} - 1/2 = 2^{n - 1}/4 + 3\pars{a_{n -1} - 1/2}$. Set
$a_{n} - 1/2 = 2^{n - 1}b_{n}/4\quad\iff\quad b_{n} = 2^{3 - n}\pars{a_{n} - 1/2}$
with $b_{3} = 1/2$.
$$
{1 \over 4}\,2^{n - 1}b_{n} = {1 \over 4}\,2^{n - 1}
+
3\,{1 \over 4}\,2^{n -2}b_{n - 1}\quad\imp\quad
b_{n} = 1 + {3 \over 2}\,b_{n - 1}
$$
$$
b_{n} + 2 = {3 \over 2}\pars{b_{n - 1} + 2}
= \pars{3 \over 2}^{2}\pars{b_{n - 2} + 2} = \cdots =
\pars{3 \over 2}^{n - 3}\bracks{b_{n - \pars{n - 3}} + 2}
$$
$$
a_{n} = \half + 2^{n - 3}b_{n}
= \half + 2^{n - 3}\bracks{-2 + \pars{3 \over 2}^{n - 3}\,{5 \over 2}}
= \half - 2^{n - 2}+ {5 \over 2}\,3^{n - 3}
$$
$$\color{#0000ff}{\large%
a_{n} = \half - 2^{n - 2}+ {5 \over 2}\,3^{n - 3}
}
$$
