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Hi, I'm actually a good student i do my own work but this stuff makes no sense to me, my teacher isn't the best so if you could provide a answer and IF possible a explanation for each problem on how you got it, it would be great. also kind of needed asap. thank you.


1. enter the explicit rule for the geometric sequence
15,3,3/5,3/25,...

an= _

2. enter a recursive rule for the geometric sequence
10,-80,640,-5120,...

a1=_; an=_

3. the reursive rule for the a geometric sequence is given
a1= 2; an= 1/3 an-1

enter the explicit rule for the sequence

an=_

4. the explicit rule for the sequence is given
an= 1/2(4/3)n-1

enter the recursive rule for the geometric sequence
a1=_; an=_

Respuesta :

LammettHash
1. Starting with 15, each successive term is obtained by multiplying by [tex]\dfrac15[/tex]. So the explicit rule for the sequence must be

[tex]a(n)=15\left(\dfrac15\right)^{n-1}[/tex]

2. Starting with 10, the next terms are obtained by multiplying by -8. So the recursive rule would be

[tex]\begin{cases}a(1)=10\\a(n)=-8a(n-1)&\text{for }n\ge2\end{cases}[/tex]

3. We're given the recursive rule,

[tex]\begin{cases}a(1)=2\\a(n)=\dfrac13a(n-1)&\text{for }n\ge2\end{cases}[/tex]

We have

[tex]a(n)=\dfrac13a(n-1)=\left(\dfrac13\right)^2a(n-2)=\left(\dfrac13\right)^3a(n-3)=\cdots=\left(\dfrac13\right)^{n-1}a(1)[/tex]

so the explicit rule is

[tex]a(n)=2\left(\dfrac13\right)^{n-1}[/tex]

4. We're given the explicit rule

[tex]a(n)=\dfrac12\left(\dfrac43\right)^{n-1}[/tex]

When [tex]n=1[/tex], we get the first term to be [tex]a(1)=\dfrac12[/tex]. For each successive term, we have to multiply by [tex]\dfrac43[/tex]. So the recursive rule is

[tex]\begin{cases}a(1)=\dfrac12\\\\a(n)=\dfrac43a(n-1)&\text{for }n\ge2\end{cases}[/tex]

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