Question

Sum of $0.4+0.44+0.444+$... $2n$ terms =

• A. $\frac{4}{81}\left(18n+1-{100}^{-n}\right)$
• B. $\frac{4}{81}\left(18n-1+{100}^{-n}\right)$
• C. $\frac{4}{81}\left(18n-1+{10}^{-n}\right)$
• D. $\frac{4}{81}\left(18n-1+{100}^n\right)$

Answer 1

The correct option is B

$\frac{4}{81}\left(18n-1+{100}^{-n}\right)$

Explanation for the correct option:

Step 1. Find the sum of AP:

$0.4+0.44+0.444+....$up to $2n$ terms

$=4\left[0.1+0.11+0.111+....\right]$

Step 2. Multiply and divide whole series by 9:

$=\frac{4}{9}\left[0.9+0.99+0.999....\right]$

$=\frac{4}{9}\left[\left(1-0.1\right)+\left(1-0.01\right)+\left(1-0.001\right)+....\right]$

$=\frac{4}{9}\left[\left(1+1+1....2nterms\right)-\left(0.1+0.01+0.001+....\right)\right]$

But $0.1,0.01,0.001,...$ are in GP

Where, $a=0.1,r=\frac{0.01}{0.1}$

$=0.1$

Step 3. Find the sum of GP:

Therefore, $S_n=\frac{4}{9}\left\{2n-0.1\left[\frac{1-{\left(0.1\right)}^{2n}}{\left(1-0.1\right)}\right]\right\}$ $\left[\mathbf{\because }{\mathit{S}}_{\mathbf{n}}\mathbf{=}\mathit{a}\frac{(1-r^n)}{\mathbf{1}\mathbf{-}\mathbf{r}}\right]$

$=\frac{4}{9}\left\{2n-\frac{0.1}{0.9}\left[1-{\left(0.1\right)}^{2n}\right]\right\}$

$=\frac{4}{9}\left\{2n-\frac{1}{9}\left[1-\frac{1}{{100}^n}\right]\right\}$

$=\frac{4}{81}\left\{18n-1+{100}^{-n}\right\}$

Hence, Option ‘B’ is Correct.