Find closed form of $a_{n+1} = 4a_n - 2$

Question

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.

Answer 1

\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.

Answer 2

We have that $-2=2/3-4\times2/3$. Therefore, $$a_{n+1}-\frac{2}{3}=4\left(a_n-\frac{2}{3}\right)$$ for all $n\in\mathbb{N}$. Therefore, $$\forall n\in\mathbb{N} : a_{n}-\frac{2}{3}=4^{n-1}\left(a_1-\frac{2}{3}\right)$$ which can be easily proven using induction.