Question

$0.2 + 0.22 + 0.222 + \ldots$ to n terms is equal to

• A. $\frac{2}{9} - \frac{2}{81}(1-10^{-n})$
• B. $\frac{n}{9} - \frac{1}{9}(1-10^{-n})$
• C. $\frac{2}{9} - \frac{1}{9}(1-10^{-n})$
• D. $\frac{2}{9}$

Hint:

$0.\overline{2} = 2/9$, $0.22 = 2/9 - 2/90$, etc.

Answer 1

The Correct Option is A

Step 1: $0.\overline{2} = \frac{2}{9}$. Each term: $0.222\ldots = \frac{2}{9}(1-10^{-k})$.}
Step 2: Sum = $\frac{2}{9}\sum_{k=1}^n (1-10^{-k}) = \frac{2}{9}\left(n - \frac{1-10^{-n}}{9}\right) = \frac{2n}{9} - \frac{2}{81}(1-10^{-n})$.}