High School

Which recursive formula can be used to define this sequence for [tex]n > 1[/tex]?

[tex]
\begin{array}{l}
84, 56, \frac{112}{3}, \frac{224}{9}, \frac{448}{27}, \frac{896}{81}, \cdots \\
A. f(n) = -\frac{4}{9} f(n-2) \\
B. f(n) = \frac{3}{2} f(n-1) \\
C. f(n) = f(n-1) - \frac{2}{9} \\
D. f(n) = \frac{2}{3} f(n-1) \\
\end{array}
[/tex]

Answer :

We are given the sequence

[tex]$$
84,\; 56,\; \frac{112}{3},\; \frac{224}{9},\; \frac{448}{27},\; \frac{896}{81},\; \dots
$$[/tex]

and we want to determine a recursive formula for the sequence when [tex]$n>1$[/tex].

Step 1. Identify the pattern between consecutive terms

First, compute the ratio between the second and the first term:

[tex]$$
\frac{56}{84} = \frac{2}{3}.
$$[/tex]

Now, check the ratio between the third and the second term:

[tex]$$
\frac{\frac{112}{3}}{56} = \frac{112}{3} \cdot \frac{1}{56} = \frac{112}{168} = \frac{2}{3}.
$$[/tex]

Since the ratio between consecutive terms is constant and equal to [tex]$\frac{2}{3}$[/tex], the sequence is a geometric sequence.

Step 2. Write the recursive formula

For a geometric sequence with a common ratio [tex]$r = \frac{2}{3}$[/tex], the recursive formula is given by

[tex]$$
f(n) = \frac{2}{3} \, f(n-1) \quad \text{for } n > 1.
$$[/tex]

Thus, the recursive formula that defines this sequence for [tex]$n>1$[/tex] is

[tex]$$
\boxed{f(n) = \frac{2}{3} \, f(n-1)}.
$$[/tex]

This is the correct answer.

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