# How do you find the nth term of the sequence 0.6, 0.61, 0.616, 0.6161,...?

Mar 29, 2017

${a}_{n} = \frac{122 - {10}^{- n} \left(77 + {\left(- 1\right)}^{n} \cdot 45\right)}{198}$

#### Explanation:

Note that:

$0. \overline{61} = \frac{61}{99}$

$0. \overline{16} = \frac{16}{99}$

So we find:

$0.6 = 0. \overline{61} - \frac{1}{10} \cdot 0. \overline{16} = \frac{61 - {10}^{- 1} \cdot 16}{99}$

$0.61 = 0. \overline{61} - \frac{1}{100} \cdot 0. \overline{61} = \frac{61 - {10}^{- 2} \cdot 61}{99}$

$0.616 = 0. \overline{61} - \frac{1}{1000} \cdot 0. \overline{16} = \frac{61 - {10}^{- 3} \cdot 16}{99}$

$0.6161 = 0. \overline{61} - \frac{1}{10000} \cdot 0. \overline{61} = \frac{61 - {10}^{- 4} \cdot 61}{99}$

. . .

We can match this alternating pattern using ${\left(- 1\right)}^{n}$...

${a}_{n} = \frac{122 - {10}^{- n} \left(77 + {\left(- 1\right)}^{n} \cdot 45\right)}{198}$