Answer:
Part A
[tex]f(n)=52-12(n-1)[/tex]
[tex]f(n)=\left\{\begin{matrix}52\: \:if \: \:n=1 & \\f(n+1)+12& if\: n\geq 2 \end{matrix}\right.[/tex]
Part B
[tex](2,4,8,16,32)\: \:[/tex] Geometric Sequence
Part C
1/4,3/4,5/4,7/4,9/4
[tex]g(n)=\frac{1}{4}+\frac{2}{4}(n-1)\\f(n)=\left\{\begin{matrix}1/4if \: \:n=1 & \\ f(n+1)+2/4& if\: n\geq 2 \end{matrix}\right[/tex]
Part D:
[tex]h(n)=1.1+0.4(n-1)\\h(n)=\left\{\begin{matrix}1.1 & if\:n=1 \\ h(n+1)+0.4 & if\:n\geq 2\end{matrix}\right[/tex]
Step-by-step explanation:
By definition, an Arithmetic Sequence holds the same difference between each following number.
Part A
[tex](52,40, 28, 16)\\52-40=12\\40-28=12\\28-16=12\\d=12[/tex]
Explicit Formula
To write an explicit formula is to write it as function.
[tex]f(n)=52-12(n-1)[/tex]
Recursive Formula
To write it as recursive formula, is to write it as recurrence given to some restrictions:
[tex]f(n)=\left\{\begin{matrix}52\: \:if \: \:n=1 & \\f(n+1)+12& if\: n\geq 2 \end{matrix}\right.[/tex]
Part B
[tex](2,4,8,16,32)\: \:[/tex]
Geometric Sequence, since 2*2=4 8*2=16 and 16*2=32 and 8+2=10 8+16=24
Part C
[tex](\frac{1}{4},\frac{3}{4},\frac{5}{4},\frac{7}{4},\frac{9}{4})\\\[/tex]
Arithmetic Sequence, difference
[tex]d=\frac{2}{4}[/tex]
Explicit Formula:
[tex]g(n)=\frac{1}{4}+\frac{2}{4}(n-1)[/tex]
Recursive Formula
[tex]g(n)=\left\{\begin{matrix}\frac{1}{4} &if\:n=1 \\ g(n+1)+\frac{2}{4} &if\: n\geq 2\end{matrix}\right.[/tex]
Part D
(1.1,1.5,1.9,2.3,2.7) Arithmetic Sequence, difference d=0.4
Explicit formula
[tex]h(n)=1.1+0.4(n-1)\\[/tex]
Recursive Formula
[tex]h(n)=\left\{\begin{matrix}1.1 &if\:n=1 \\ h(n+1)+0.4 &if\: n\geq 2\end{matrix}\right.[/tex]
