Look at the decimal 0.121122111222... If the pattern continues, is this a repeating decimal?

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## 3 Answers

NO it DONT repeet...number av "2" vary
by
is 0.121122111222... a repeating decimal if it kepps going
by

This is not a recurring decimal even though it is a repeating pattern.

This decimal can be represented by the sum of two series. First write the number as 0.10110011100011110000... + 0.0200220002220000...

We can write this sum as:

(⅟₉)(0.9099009990009999... + 2×(0.090099000999...)).

We can write this as:

(⅟₉)∑[(10ⁿ-1)/10^n²)(1+2/10ⁿ)] where ∑ means “sum of” and is applied to the terms in the square brackets where n is an integer greater than 0.

From this we can see that the nth term is expressed by the quantity in the square brackets. To see this more clearly put n=1 to give us the first term:

(⅟₉)(10-1)/10)(1.2)=0.12=3/25.

Now, n=2:

(⅟₉)(100-1)/10000)(1.02)=0.001122=561/500000;

and n=3:

(⅟₉)(1000-1)/10⁹)(1.002)=0.000000111222=55611/(5×10¹¹).

Add these together: 0.121122111222, a finite decimal.

A recurring decimal always represents one unique fraction, but it’s clear that the fraction given by the formula produces a finite decimal (with n²+n decimal places) and a changing fraction as n increases. Therefore, the given decimal cannot be a repeating (recurring) decimal and it has an infinite number of decimal places as n➝∞.

by Top Rated User (639k points)

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