# A geometric sequence is defined recursively by a_n = 5a_(n-1), the first term of the sequence is 0.45. What is the explicit formula for the nth?

Mar 5, 2016

$0.45 \times {5}^{n - 1}$

#### Explanation:

In a geometric series $\left\{a , a r , a {r}^{2} , a {r}^{3} , \ldots . .\right\}$, $a$ is first term and ratio is $r$, ${n}^{t h}$ term is given by $a {r}^{n - 1}$. Note $r = {a}_{m} / {a}_{m - 1}$ or ${a}_{m} = r {a}_{m - 1}$

As ${a}_{n} = 5 {a}_{n - 1}$, $r = 5$ and first term $a = 0.45$

Hence, ${n}^{t h}$ term is given by $0.45 \times {5}^{n - 1}$.

Mar 5, 2016

${a}_{n} = {5}^{\left(n - 1\right)} \left(0.45\right)$

#### Explanation:

The recursive definition ${a}_{n} = 5 {a}_{n - 1}$ means that

${a}_{2} = 5 {a}_{1}$

${a}_{\textcolor{red}{3}} = 5 {a}_{2}$

$= 5 \cdot \left(5 {a}_{1}\right)$

$= {5}^{2} {a}_{1}$

$= {5}^{\left(\textcolor{red}{3} - 1\right)} {a}_{1}$

${a}_{\textcolor{red}{4}} = 5 {a}_{3}$

$= 5 \cdot \left({5}^{2} {a}_{1}\right)$

$= {5}^{3} {a}_{1}$

$= {5}^{\left(\textcolor{red}{4} - 1\right)} {a}_{1}$

$\ldots$

We can guess that

${a}_{\textcolor{red}{n}} = {5}^{\left(\textcolor{red}{n} - 1\right)} {a}_{1}$

and show that this is true by induction.

For $n = 1$,

${a}_{1} = {5}^{\left(1 - 1\right)} {a}_{1} = {5}^{0} {a}_{1} = {a}_{1}$.

For all positive integers $n$ greater than 1,

${a}_{n} = {5}^{\left(n - 1\right)} {a}_{1}$

$= 5 \cdot {5}^{\left(n - 2\right)} {a}_{1}$

$= 5 \cdot \left({5}^{\left(\left(n - 1\right) - 1\right)} {a}_{1}\right)$

$= 5 {a}_{n - 1}$.

Since it is given that the first term (${a}_{1}$) is 0.45, the explicit formula for the ${n}^{\text{th}}$ term is

${a}_{n} = {5}^{\left(n - 1\right)} \left(0.45\right)$.