Question

A sequence has the recursive definition of f(1)=2/3, f(n)=f(n-1)×3/2 for n≥2


Right the first five terms of the sequence

Leave answers as fractions

Answer 1

Answer:

Step-by-step explanation:

$f(1)=2/3\\f(2)=f(1)\times (3/2)=(2/3)\times(3/2)=1\\f(3)=f(2)\times (3/2) = 1\times (3/2) = 3/2\\f(4)=f(3)\times (3/2) = (3/2)\times (3/2) = 9/4\\f(5)=(f(4)\times (3/2) = (9/4)\times (3/2) = 27/8$

So the first five terms are 2/3, 1, 3/2, 9/4, 27/8.