Question

Which sequence can be defined by the recursive formula f (1) = 4, f (n + 1) = f (n) – 1.25 for n ≥ 1? 1, –0.25, –1.5, –2.75, –4, . . . 1, 2.25, 3.5, 4.75, 6, . . . 4, 2.75, 1.5, 0.25, –1, . . . 4, 5.25, 6.5, 7.75, 8, . . .

Answer 1

Answer:

$4,2.75,1.5,0.25......$

Step-by-step explanation:

f (1) = 4, $f (n + 1) = f (n) - 1.25$

To get the sequence we start with n=1

Plug in n=1 in the formula

$f (n + 1) = f (n) - 1.25$

$f (1 + 1) = f (1) - 1.25$

$f (2) = f (1) - 1.25$, replace f(1)=4

$f (2) = 4 - 1.25=2.75$

n=2

$f (2 + 1) = f (2) - 1.25$

$f (3) = f (2) - 1.25$, replace f(2)=2.75

$f (3) = 2.75 - 1.25=1.5$

n=3

$f (3 + 1) = f (3) - 1.25$

$f (4) = f (3) - 1.25$, replace f(3)=1.5

$f (4) = 1.5 - 1.25=0.25$

Sequence is $4,2.75,1.5,0.25......$

Answer 2

send picture of problem ?

Answer 3

the answer is C)