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Math Problems
Algebra 1
Evaluate recursive formulas for sequences
{
g
(
1
)
=
4
g
(
2
)
=
−
3
g
(
n
)
=
g
(
n
−
2
)
⋅
g
(
n
−
1
)
g
(
3
)
=
\begin{array}{l}\left\{\begin{array}{l}g(1)=4 \\ g(2)=-3 \\ g(n)=g(n-2) \cdot g(n-1)\end{array}\right. \\ g(3)=\end{array}
⎩
⎨
⎧
g
(
1
)
=
4
g
(
2
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=
−
3
g
(
n
)
=
g
(
n
−
2
)
⋅
g
(
n
−
1
)
g
(
3
)
=
Get tutor help
{
g
(
1
)
=
−
5
g
(
2
)
=
3
g
(
n
)
=
g
(
n
−
2
)
+
g
(
n
−
1
)
g
(
3
)
=
□
\begin{array}{l}\left\{\begin{array}{l}g(1)=-5 \\ g(2)=3 \\ g(n)=g(n-2)+g(n-1)\end{array}\right. \\ g(3)=\square\end{array}
⎩
⎨
⎧
g
(
1
)
=
−
5
g
(
2
)
=
3
g
(
n
)
=
g
(
n
−
2
)
+
g
(
n
−
1
)
g
(
3
)
=
□
Get tutor help
{
f
(
1
)
=
−
2
f
(
2
)
=
5
f
(
n
)
=
f
(
n
−
2
)
⋅
f
(
n
−
1
)
f
(
3
)
=
\begin{array}{l}\left\{\begin{array}{l}f(1)=-2 \\ f(2)=5 \\ f(n)=f(n-2) \cdot f(n-1)\end{array}\right. \\ f(3)=\end{array}
⎩
⎨
⎧
f
(
1
)
=
−
2
f
(
2
)
=
5
f
(
n
)
=
f
(
n
−
2
)
⋅
f
(
n
−
1
)
f
(
3
)
=
Get tutor help
{
f
(
1
)
=
−
6
f
(
2
)
=
−
4
f
(
n
)
=
f
(
n
−
2
)
+
f
(
n
−
1
)
f
(
3
)
=
\begin{array}{l}\left\{\begin{array}{l}f(1)=-6 \\ f(2)=-4 \\ f(n)=f(n-2)+f(n-1)\end{array}\right. \\ f(3)=\end{array}
⎩
⎨
⎧
f
(
1
)
=
−
6
f
(
2
)
=
−
4
f
(
n
)
=
f
(
n
−
2
)
+
f
(
n
−
1
)
f
(
3
)
=
Get tutor help
{
f
(
1
)
=
15
f
(
n
)
=
f
(
n
−
1
)
⋅
n
f
(
2
)
=
□
\begin{array}{l}\left\{\begin{array}{l}f(1)=15 \\ f(n)=f(n-1) \cdot n\end{array}\right. \\ f(2)=\square\end{array}
{
f
(
1
)
=
15
f
(
n
)
=
f
(
n
−
1
)
⋅
n
f
(
2
)
=
□
Get tutor help
{
g
(
1
)
=
0
g
(
n
)
=
g
(
n
−
1
)
+
n
g
(
2
)
=
□
\begin{array}{l}\left\{\begin{array}{l}g(1)=0 \\ g(n)=g(n-1)+n\end{array}\right. \\ g(2)=\square\end{array}
{
g
(
1
)
=
0
g
(
n
)
=
g
(
n
−
1
)
+
n
g
(
2
)
=
□
Get tutor help
{
f
(
1
)
=
1
f
(
2
)
=
2
f
(
n
)
=
f
(
n
−
2
)
+
f
(
n
−
1
)
f
(
3
)
=
□
\begin{array}{l}\left\{\begin{array}{l}f(1)=1 \\ f(2)=2 \\ f(n)=f(n-2)+f(n-1)\end{array}\right. \\ f(3)=\square\end{array}
⎩
⎨
⎧
f
(
1
)
=
1
f
(
2
)
=
2
f
(
n
)
=
f
(
n
−
2
)
+
f
(
n
−
1
)
f
(
3
)
=
□
Get tutor help
{
h
(
1
)
=
−
35
h
(
n
)
=
h
(
n
−
1
)
⋅
2
h
(
3
)
=
\begin{array}{l}\left\{\begin{array}{l}h(1)=-35 \\ h(n)=h(n-1) \cdot 2\end{array}\right. \\ h(3)=\end{array}
{
h
(
1
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=
−
35
h
(
n
)
=
h
(
n
−
1
)
⋅
2
h
(
3
)
=
Get tutor help
Find
a
2
a_2
a
2
and
a
3
a_3
a
3
.
\newline
\newline
a
1
=
2
a_1 = 2
a
1
=
2
\newline
a
n
=
2
a
n
−
1
a_n = 2a_{n - 1}
a
n
=
2
a
n
−
1
\newline
\newline
Write your answers as integers or fractions in simplest form.
\newline
\newline
a
2
=
a_2 =
a
2
=
______
\newline
a
3
=
a_3 =
a
3
=
______
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