Question

A recursive rule for a geometric sequence is a1=2; an= 4/9 an-1

Answer 1

A recursive rule for a geometric sequence:

$a_1\\\\a_n=r\cdot a_{n-1}$

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$a_1=2;\ a_n=\dfrac{4}{9}a_{n-1}\to r=\dfrac{4}{9}$

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The exciplit rule:

$a_n=a_1r^{n-1}$

Substitute:

$a_n=2\left(\dfrac{4}{9}\right)^{n-1}=2\cdot\left(\dfrac{4}{9}\right)^n\cdot\left(\dfrac{4}{9}\right)^{-1}=2\cdot\left(\dfrac{4}{9}\right)^n\cdot\dfrac{9}{4}\\\\\boxed{a_n=\dfrac{9}{2}\cdot\left(\dfrac{4}{9}\right)^n}$