Chapter 7.8, Problem 29PPE

To determine

Tofind:The explicit formula for thegiven geometric sequence.

Answer to Problem 29PPE

Thus, the recursive formula for the geometric sequence $1,5,25,125,\dots$ is $a_1=1$ , and $a_{n+1}=5\cdot a_n,n\ge 1$ .

Explanation of Solution

Given information:

The given geometric sequence is $1,5,25,125,\dots$ .

Conceptused:

A geometric sequence is a sequence in which the ratio consecutive terms is constant.

A sequence of the form $a_1,a_{1}r,a_1{r}^2,a_1{r}^3,\dots ,a_1{r}^{n-1},\dots$ is the geometric with $a_1$ as the first term, r the common ratio, and nth term $a_n=a_1{r}^{n-1}$ .This is explicit formula.

The recursive formula for the geometric sequence is the iterative relation between consecutive terms ${\left(n+1\right)}^{\text{th}}$ and n^th terms that is given by $a_{n+1}=r\cdot a_n,n\ge 1$

The given geometric sequence is $\langle s\rangle =1,5,25,125,\dots$ , that may be written as

  $\langle s\rangle =1,1\times \left(5\right),1\times {\left(5\right)}^2,1\times {\left(5\right)}^3,\dots ,1\times {\left(5\right)}^{n-1}=a_n,1\times {\left(5\right)}^n=a_{n+1},\dots$ .

The first term of the sequence is1and the common ratio is5.

The n^th term of sequence is $a_n=1\times {\left(5\right)}^{n-1}$ and the ${\left(n+1\right)}^{\text{th}}$ term $a_{n+1}=1\times {\left(5\right)}^n$ .

Now, $\frac{a_{n+1}}{a_n}=\frac{1\times {\left(5\right)}^n}{1\times {\left(5\right)}^{n-1}}=5$

  $\Rightarrow a_{n+1}=5\times a_n$ .

Thus, the recursive formula for the geometric sequence $1,5,25,125,\dots$ is $a_1=1$ , and $a_{n+1}=5\cdot a_n,n\ge 1$ .

Conclusion:

Thus, the recursive formula for the geometric sequence $1,5,25,125,\dots$ is $a_1=1$ , and $a_{n+1}=5\cdot a_n,n\ge 1$ .