# How do you write an explicit formula for the sequence 40, 20, 10, 5, 2.5...?

Mar 28, 2016

Explicit formula for the ${n}^{t h}$ term ${T}_{n}$ of the series is $40 {\left(\frac{1}{2}\right)}^{n - 1}$ and sum of the series $S$ is given by $80 \left(1 - \frac{1}{{2}^{n}}\right)$

#### Explanation:

In the series $\left\{40 , 20 , 10 , 5 , 2.5 \ldots\right\}$, the ratio of a term to its preceding term is $\frac{20}{40} = \frac{10}{20} = \frac{5}{10} = \frac{2.5}{5} = \ldots = \frac{1}{2}$ always constant. Hence it is a geometric series of type $\left\{a , a r , a {r}^{2} , a {r}^{3} , \ldots .\right\}$ with first term $a = 40$ and $r = \frac{1}{2}$.

As is apparent ${n}^{t h}$ term of the series

${T}_{n} = a {r}^{n - 1}$ or $40 {\left(\frac{1}{2}\right)}^{n - 1}$.

Let the sum of the series be $S$, then

$S = a + a r + a {r}^{2} + a {r}^{3} + \ldots . + a {r}^{n - 1}$ -------(A)

and multiplying both sides by $r$, we get

$r S = a r + a {r}^{2} + a {r}^{3} + a {r}^{4.} \ldots + a {r}^{n}$ -------(B)

Subtracting (B) from (A)

$\left(1 - r\right) S = a {r}^{n} - a = a \left(1 - {r}^{n}\right)$

Hence $S = \frac{a \left(1 - {r}^{n}\right)}{1 - r} = 40 \frac{1 - \frac{1}{{2}^{n}}}{1 - \frac{1}{2}} = 40 \frac{1 - \frac{1}{{2}^{n}}}{\frac{1}{2}} = 80 \left(1 - \frac{1}{{2}^{n}}\right)$