<s>[INST] Find closed form of $a_{n+1} = 4a_n - 2$ Is there a easy way to find the recurrence for $a_{n+1} = 4a_n - 2$, with $a_1 = 2$?
I was learning about finding the closed form of Fibonacci-styled recurrences like $a_{n+2} = 2a_{n+1} - 3a_{n}$ or something, and then this problem appeared. The textbook did a ton of rearranging to cancel out the $-2$ term and get it into the form of a Fibonacci-styled recurrence. I'm wondering if there is a easier way.
 [/INST] \begin{align*}
a_2 &= 4a_1 - 2\\
a_3 &= 4(4a_1-2)-2 = 4^2a_1 - 4\cdot 2 - 2\\
a_4 &= 4(4^2a_1 - 4\cdot 2 - 2) - 2 = 4^3a_1 - 4^2\cdot 2 - 4\cdot 2 - 2.
\end{align*}
This leads to the conjecture
$$
a_n = 4^{n-1}a_1 - 2\sum_{k=0}^{n-2}4^k = 4^{n-1}a_1 - 2\frac{4^{n-1}-1}{4-1} = 4^{n-1}(a_1-\tfrac 23) + \tfrac 23.
$$
It is easily checked that this sequence satisfies the iteration rule, hence it's the solution.
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