High School

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

[tex]44, -11, \frac{11}{4}, -\frac{11}{16}, \frac{11}{64}, -\frac{11}{256}, \ldots[/tex]

A. [tex]f(n) = f(n-1) - 55[/tex]

B. [tex]f(n) = -4 f(n-1)[/tex]

C. [tex]f(n) = -\frac{1}{4} f(n-1)[/tex]

D. [tex]f(n) = \frac{1}{16} f(n-2)[/tex]

Answer :

We begin by writing the sequence:

[tex]$$
a_1 = 44,\quad a_2 = -11,\quad a_3 = \frac{11}{4},\quad a_4 = -\frac{11}{16},\quad \ldots
$$[/tex]

Step 1: Calculate the ratio between the second term and the first term:

[tex]$$
\text{ratio} = \frac{a_2}{a_1} = \frac{-11}{44} = -\frac{1}{4}.
$$[/tex]

Step 2: Verify that this common ratio is consistent by computing the third term:

[tex]$$
a_3 = a_2 \times \left(-\frac{1}{4}\right) = -11 \times \left(-\frac{1}{4}\right) = \frac{11}{4}.
$$[/tex]

Since the ratio remains [tex]$-\frac{1}{4}$[/tex] from term to term, the sequence is geometric.

Step 3: Therefore, the recursive formula for the sequence is:

[tex]$$
f(n) = -\frac{1}{4} \, f(n-1) \quad \text{for } n > 1.
$$[/tex]

Among the options provided, this corresponds to the third option.