Given the sequence series $$ \{a_n\} $$ satisfies $$ a_1=2 $$ , $$ a_n=3a_{n-1}+1 $$ if $$ n≥2 $$ . If $$ b_n = a_n+\dfrac{1}{2} $$
(1) prove $$ \{b_n\} $$ is a geometric sequence
(2) Find the general term formula of $$ \{a_n\} $$

Question

Given the sequence series $$ \{a_n\} $$ satisfies $$ a_1=2 $$ , $$ a_n=3a_{n-1}+1  $$ if $$ n≥2 $$ . If $$ b_n = a_n+\dfrac{1}{2}  $$
(1) prove  $$ \{b_n\} $$ is a geometric sequence
(2) Find the general term formula of  $$ \{a_n\} $$

Uniteasy Admin · Accepted Answer

(1)
$$ \because b_n = a_n+\dfrac{1}{2} $$
 $$ 	herefore b_{n+1} = a_{n+1}+\dfrac{1}{2} $$ 
$$ \dfrac{ b_{n+1}}{ b_n} =\dfrac{ a_{n+1}+\dfrac{1}{2}}{a_n+\dfrac{1}{2} } 
 $$
$$ \quad\quad\space\space =\dfrac{3a_n+1+\dfrac{1}{2}}{a_n+\dfrac{1}{2}} $$
$$ \quad\quad\space\space =3 $$
$$ 	herefore  \{b_n\} $$ is a geometric sequence, common ratio is 3
(2) 
$$ b_1 = a_1+\dfrac{1}{2} = \dfrac{5}{2} $$
$$ \because  \{b_n\} $$ is a geometric sequence, common ratio is 3
$$ b_n = a_1q^{n-1} = \dfrac{5}{2} \cdotp 3^{n-1} = \dfrac{5}{6}3^n $$
$$ a_n = b_n-\dfrac{1}{2} $$
$$ \quad \space\space =\dfrac{5}{6}3^n -\dfrac{1}{2} $$

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