Given the recursively defined formula a1=β3
π
1
=
β
3
, an=anβ1+4
π
π
=
π
π
β
1
+
4
, determine which option is the correct function notation.(1 point)
Responses
a(1)=β3
π
(
1
)
=
β
3
; f(n)=f(nβ1)+4
π
(
π
)
=
π
(
π
β
1
)
+
4
a left parenthesis 1 right parenthesis equals negative 3 ; f left parenthesis n right parenthesis equals f left parenthesis n minus 1 right parenthesis plus 4
f(1)=β3
π
(
1
)
=
β
3
; f(n)=f(anβ1)+4
π
(
π
)
=
π
(
π
π
β
1
)
+
4
f left parenthesis 1 right parenthesis equals negative 3 ; f left parenthesis n right parenthesis equals f left parenthesis a subscript n minus 1 baseline right parenthesis plus 4
f(1)=β3
π
(
1
)
=
β
3
; f(n)=f(nβ1)+4
π
(
π
)
=
π
(
π
β
1
)
+
4
f left parenthesis 1 right parenthesis equals negative 3 ; f left parenthesis n right parenthesis equals f left parenthesis n minus 1 right parenthesis plus 4
f(1)=β3
π
(
1
)
π
1
=
β
3
, an=anβ1+4
π
π
=
π
π
β
1
+
4
, determine which option is the correct function notation.(1 point)
Responses
a(1)=β3
π
(
1
)
=
β
3
; f(n)=f(nβ1)+4
π
(
π
)
=
π
(
π
β
1
)
+
4
a left parenthesis 1 right parenthesis equals negative 3 ; f left parenthesis n right parenthesis equals f left parenthesis n minus 1 right parenthesis plus 4
f(1)=β3
π
(
1
)
=
β
3
; f(n)=f(anβ1)+4
π
(
π
)
=
π
(
π
π
β
1
)
+
4
f left parenthesis 1 right parenthesis equals negative 3 ; f left parenthesis n right parenthesis equals f left parenthesis a subscript n minus 1 baseline right parenthesis plus 4
f(1)=β3
π
(
1
)
=
β
3
; f(n)=f(nβ1)+4
π
(
π
)
=
π
(
π
β
1
)
+
4
f left parenthesis 1 right parenthesis equals negative 3 ; f left parenthesis n right parenthesis equals f left parenthesis n minus 1 right parenthesis plus 4
f(1)=β3
π
(
1
)
Answers
Answered by
GPT-4o mini
The correct function notation that corresponds to the recursively defined formula \( a_1 = -3 \) and \( a_n = a_{n-1} + 4 \) is:
**f(1) = -3; f(n) = f(n-1) + 4**
This captures the initial condition and the recursive relationship accurately. Thus, the answer is:
**f(1)=β3; f(n)=f(nβ1)+4**
**f(1) = -3; f(n) = f(n-1) + 4**
This captures the initial condition and the recursive relationship accurately. Thus, the answer is:
**f(1)=β3; f(n)=f(nβ1)+4**