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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.

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2 Answers 2

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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 $a_n-\dfrac12=\dfrac52\times 3^n$.

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  • $\begingroup$ I'm not sure of understanding how the formulae is obtained... $\endgroup$
    – user108343
    Commented Sep 20, 2015 at 6:13
  • $\begingroup$ @Astroman Set $u_n=a_n-\dfrac12$. From $u_n=3u_{n-1}$, $u_0=5/2$ (this is a geometric sequence) you easily get $u_n=\dfrac52\times 3^n$ giving the formula$$ a_n=\dfrac52\times 3^n+\frac12.$$ $\endgroup$
    – Olivier Oloa
    Commented Sep 20, 2015 at 6:18
  • $\begingroup$ I have two main questions : how did you know what to replace by what ? I mean, how did you get the 1/2 ? how am I supposed to know that ? (I have never done series ) Also, when you do replace all of the stuff, how did you solve for u0=5/2 ? (Im new to this) $\endgroup$
    – user108343
    Commented Sep 20, 2015 at 6:25
  • $\begingroup$ Ok, I understand how you solved for u0, but I dn'T know how you figured all of this. Where does the 1/2 come from ? $\endgroup$
    – user108343
    Commented Sep 20, 2015 at 6:32
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    $\begingroup$ While Im not completely sure why you think its not necessary. (For me, this would make it be an inequality) Im going to take it this way anyway. Ok thanks. ! !! $\endgroup$
    – user108343
    Commented Sep 20, 2015 at 16:21
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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 $b(n)=3b(n-1)$

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