Which sequence could be partially defined by the recursive formula f (n + 1) = f(n) + 2.5 for n ≥ 1?

Options:

a. 2.5, 6.25, 15.625, 39.0625, …

b. 2.5, 5, 10, 20

c. -10, -7.5, -5, -2.5, …

d. -10, -25, 62.5, 156.25

Answer:

C. –10, –7.5, –5, –2.5, …

Step-by-step explanation:

Given

f(n+1) = f(n) + 2.5 where n ≥ 1

Required

Which sequence partially defines the function

This question will be tackled using a trial and error approach (we’ll evaluate each option provided)

Let’s consider f(1) as the starting point for each sequence since n >= 1.

A.

2.5, 6.25, 15.625, 39.0625, …

f(1) = 2.5

f(n+1) = f(n) + 2.5

So when n = 1

f(2) = f(1) + 2.5 = 5

This option does not define the sequence.

B.

2.5, 5, 10, 20

f(1) = 2.5

So when n = 1

f(2) = f(1) + 2.5 = 5

For n = 2

f(3) = f(2) + 2.5 = 7.5

This option also does not define the sequence.

C.

-10, -7.5, -5, -2.5, …

f(1) = -10

So when n = 1

f(2) = f(1) + 2.5 = -7.5

For n = 2

f(3) = f(2) + 2.5 = -5

For n = 3

f(4) = f(3) + 2.5 = -2.5

This defines the function, so there’s no need to assess D.