Finding a closed form for the recursive sequence $a_n=\frac13(10-2a_{n-1})$, with $a_0=3$

Question

I've got following recursive sequence: $$ a_n=\frac13(10-2a_{n-1}) $$ With $ a_0=3$

I have found following expression:

$$ a_n=3+ \sum_{n=0}^n{2^{n-1}\cdot\frac{5}{3^n}\cdot (-1)^n} $$

Is there maybe a better form? Thank you for your help! I would appreciate hints

Answer 1

Let $\,b_n=a_n-2\;\;$ for any $\,n\in\Bbb N\cup\{0\}\,.\;\;\left(\implies\color{brown}{b_0=1}\,\right)$

From $\,a_n=\dfrac{10-2a_{n-1}}3\;\;\forall n\in\Bbb N\;,\;$ we get that

$b_n+2=\dfrac{10-2\left(b_{n-1}+2\right)}3\;\;\;\forall n\in\Bbb N\;,$

$b_n=\dfrac{6-2b_{n-1}}3-2\;\;\;\forall n\in\Bbb N\;,$

$b_n=-\dfrac23b_{n-1}\;\;\;\forall n\in\Bbb N\;.$

Consequently ,

$b_n=-\dfrac23b_{n-1}=\left(-\dfrac23\right)^{\!2}\!b_{n-2}=\ldots=\left(-\dfrac23\right)^{\!n}\!b_0=\left(-\dfrac23\right)^{\!n}\;\;,$

$a_n=2+b_n=2+\left(-\dfrac23\right)^{\!n}\;\;$ for any $\,n\in\Bbb N\cup\{0\}\,.$