Question

Sum the series: $.4+.44+.444+$_____ to $n$ terms.

Answer 1

Now,

$.4+.44+.444+$_____ to $n$ terms

$=\frac{4}{9}(.9+.99+.999+$_____ to $n$ terms$)$

$=\frac{4}{9}\left\{\right(1-.1)+(1-.01)+(1-.001)+$_____ to $n$ terms$\}$

$=\frac{4}{9}\{n-(\frac{1}{10}+\frac{1}{{10}^2}+......+\frac{1}{{10}^n})\}$

$=\frac{4}{9}\{n-\left(\frac{1-\frac{1}{{10}^n}}{1-\frac{1}{10}}\right)\}$

$=\frac{4}{9}\{n-\frac{10}{9}\left(1-\frac{1}{{10}^n}\right)\}$.