Question

Find the sum of the sequence $0.4+0.44+0.444+...$ up to $n$ terms.

Answer 1

Finding the sum of $0.4+0.44+0.444+...$

Let,

$$\begin{gathered} S=0.4+0.44+0.444+...nterms \\ \Rightarrow S=4\left(0.1+0.11+0.111+...nterms\right) \end{gathered}$$

Multiplying and dividing by $9$, we get,

$$\begin{gathered} \Rightarrow S=4\left(0.1+0.11+0.111+...nterms\right) \\ \Rightarrow S=\frac{4}{9}(0.9+0.99+0.999+...nterms) \\ \Rightarrow S=\frac{4}{9}\left[\left(1-0.1\right)+\left(1-0.01\right)+\left(1-0.001\right)+...nterms\right] \\ \Rightarrow S=\frac{4}{9}\left[\left(1+1+1+...nterms\right)-\left(0.1+0.01+0.001+...+{\left(0.1\right)}^n\right)\right] \\ \Rightarrow S=\frac{4}{9}\left[n-\frac{0.1\left(1-0.1^n\right)}{1-0.1}\right] \\ \Rightarrow S=\frac{4}{9}\left[n-\frac{1-{\left({\displaystyle \frac{1}{10}}\right)}^n}{9}\right] \\ \Rightarrow S=\frac{4}{9}\left[9n-1+\frac{1}{{10}^n}\right] \end{gathered}$$

Therefore, the sum of $0.4+0.44+0.444+...$ up to $n$ terms. is $\frac{4}{9}\left[9n-1+\frac{1}{{10}^n}\right]$