Question

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

Answer 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 solved using trial by error method (we'll check out all given options)

We'll take f(1) as the first time of each sequence because 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(1 + 1) = f(1) + 2.5

f(2) = 2.5 + 2.5 = 5

This doesn't define the sequence

B.

2.5, 5, 10, 20

f(1) = 2.5

So, when n = 1

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

f(2) = 2.5 + 2.5 = 5

When n = 2

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

f(3) = 5 + 2.5

f(3) = 7.5

This also doesn't define the sequence

C.

-10, -7.5, -5, -2.5, …

f(1) = -10

So, when n = 1

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

f(2) = -10 + 2.5 = -7.5

When n = 2

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

f(3) = -7.5 + 2.5 = -5

When n = 3

f(3 + 1) = f(3) + 2.5

f(4) = -5 + 2.5 = -2.5

This defines the function; so, there's no need to check D