Explained: "(a) write the repeating decimal as a geometric..."

boitshupoO

Answered question

2020-12-31

(a) write the repeating decimal as a geometric series, and (b) write its sum as the ratio of two integers.

0. \overset{―}{81}

Nathanael Webber

Skilled2021-01-01Added 117 answers

Given decimal:

0. \overset{―}{81}

(a) For the repeating decimal 0.81, we can write

0. \overset{―}{81} = 0.818181 . . .

= 0.81 + 0.0081 + 0.000081 + 0.00000081 + . . .

= \frac{81}{10^{2}} + \frac{81}{10^{4}} + \frac{81}{10^{6}} + . . .

= \sum_{n = 0}^{\infty} \frac{81}{10^{2}} (\frac{1}{10^{2}})^{n}

(b) Given Series:

= \sum_{n = 0}^{\infty} \frac{81}{10^{2}} (\frac{1}{10^{2}})^{n}

Given series is a Geometric series with ratio

r = \frac{1}{10^{2}}

and

a_{1} = \frac{81}{10^{2}}

So the sum is

= \frac{a_{1}}{(1 - r)}

= \frac{\frac{81}{100}}{(1 - \frac{1}{100})}

= \frac{81}{99}

= \frac{9}{11}

Result

\frac{9}{11}

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