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The sum of 0.2 +0.22+0.222+ … to n ter...

The sum of ` 0.2 +0.22+0.222+ …` to n terms is equal to

A

`((2)/(9))-((2)/(81))(1-10^(-n))`

B

`n((1)/(9))(1-10^(-n))`

C

`((2)/(9))[n-((1)/(9))(1-10^(-n))]`

D

`((2)/(9))`

Text Solution

Verified by Experts

The correct Answer is:
C
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