Question

The recursive rule is a1=__and a n-1 for n> __.the explicit rule is a n= ___(__)n-1

Answer 1

We are required to find both the recursive rule and the Explicit rule for the values given in the table.

Recursive rule:

To get the recursive rule, we need to find the common ratio.

The common ratio (r) is found by dividing the 2nd term by the 1st term, or the 3rd term by the 2nd term, or the 4th term by the 3rd term, and so on.

We can generalize and say the common ratio is also the division of the nth term by the (n-1)th term.

Let us put these facts into mathematical expressions:

`\begin{gathered} Let,_{}_{} \\ a_1=\text{first term} \\ a_2=\text{Second term} \\ a_3=\text{third term} \\ \text{...} \\ a_n=\text{nth term} \\ \\ \text{common ratio (r)=}(a_1)/(a_2)=(a_3)/(a_2)=(a_4)/(a_2)\ldots=(a_n)/(a_(n-1)) \\ \\ \text{if a}_1=1,a_2=3_{} \\ \therefore r=(a_2)/(a_1)=(3)/(1)=3 \end{gathered}`

Thus, we can now write the recursive formula as:

`\begin{gathered} (a_n)/(a_(n-1))=\text{common ratio (r)} \\ \\ \therefore(a_n)/(a_(n-1))=3 \\ \\ \therefore a_1=1,a_n=3a_(n-1)\text{ for n}\ge1 \end{gathered}`

Explicit rule:

To get the rule, we need to recognize the pattern:

`\begin{gathered} \text{when n = 1} \\ a_1=1=3^0 \\ \\ \text{When n = 2} \\ a_2=3=3^1 \\ \\ \text{When n= 3} \\ a_3=9=3^2 \\ \\ \text{When n= 4} \\ a_4=27=3^3 \\ \\ \text{When n = 5} \\ a_5=81=3^4 \end{gathered}`

We can see a pattern developing and from that, we can generalize:

`\begin{gathered} a_1=3^0 \\ a_2=3^1 \\ a_3=3^2 \\ a_4=3^3 \\ a_5=3^4 \\ \ldots \\ a_n=3^(n-1) \end{gathered}`

Thus, the explicit rule is:

`\begin{gathered} a_1=1 \\ a_n=1*3^(n-1) \end{gathered}`