Finding the general formula of a sequence: $3,8,23,68,203,608,\cdots$

Question

I have the following sequence : $$3,8,23,68,203,608,\cdots$$

I have found that definition by recurrence of this is $$a(n)=3a(n-1)-1$$ where $a_0=3$ as the first term.

I want to find the explicit formula for this sequence but I have no idea how to find it.

Answer 1

Hint. You may observe that

$$ a_n-\frac12=3\left(a_{n-1}-\frac12\right),\qquad a_0=3, $$

then you may easily obtain an explicit formula for $a_n-\dfrac12$ thus an explicit formula for $a_n$.

Addendum. From $a_n=3a_{n-1}-1$, we are looking for a real number $\alpha$ such that $$ \color{red}{a_n}+\alpha=3\left(a_{n-1}+\alpha\right) \tag1 $$ then replacing $\color{red}{a_n}$ with $\color{red}{3a_{n-1}-1}$ and expanding $(1)$ gives $$ \color{red}{3a_{n-1}-1}+\alpha=3a_{n-1}+3\alpha \tag2 $$ and $(2)$ rewrites $ -1+\alpha=3\alpha$ from which you deduce $\alpha=-\dfrac12$. The first term of the auxiliary sequence is $a_0-\dfrac12=3-\dfrac12=\dfrac52$. Thus from $(1)$ you get $a_n-\dfrac12=\dfrac52\times 3^n$.

Answer 2

HINT:

Let $a(n)= b(n)+cn+d$

$$b(n)+cn+d=3\{b(n-1)+c(n-1)+d\}-1$$

$$\iff b(n)+cn+d=3b(n-1)+3cn+3d-3c-1$$

Set $c=0,d=3d-3c-1$ to get $b(n)=3b(n-1)$