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\}

Steven Zheng Steven Zheng posted 2 months ago


Answer

(1)

\because b_n = a_n+\dfrac{1}{2}

\therefore 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

\therefore \{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}

Steven Zheng Steven Zheng posted 2 months ago

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