Question

Find the sum to n terms: .4 + .44 + .444​

Answer 1

$Given \: \: Question \: \: Is \: \\ \\ 0.4 + 0.44 + 0.444 + ... + n \\ \\ Answer \: \: \\ \\ let \: \: \ \\ s = 0.4 + 0.44 + 0.444 + 0.444 + ... \\ the \: above \: equation \: can \: be \: re \: write \: as \\ \\ s = \frac{4}{10} + \frac{44}{100} + \frac{444}{1000} + \frac{4444}{10000} + ... \\ \\ s = 4( \frac{1}{10} + \frac{11}{100} + \frac{111}{1000} + \frac{1111}{10000} + ...) \\ \\ s = \frac{4}{9} ( \frac{9}{10} + \frac{99}{100} + \frac{999}{1000} + \frac{9999}{10000} + ...) \\ \\ s = \frac{4}{9} ((1 - \frac{1}{10} ) + (1 - \frac{1}{100} ) + (1 - \frac{1}{1000} ) + (1 - \frac{1}{10000} ) + ...) \\ \\ s = \frac{4}{9} (1 + 1 + 1 + 1 + ...) - \frac{4}{9} ( \frac{1}{10} + \frac{1}{100} + \frac{1}{1000} + \frac{1}{10000} + ...) \\ \\ s = \frac{4n}{9} - \frac{4}{90} ( 1 + \frac{1}{10} + \frac{1}{100} + \frac{1}{1000} + ...) \\ \\ s = \frac{4n}{9} - \frac{2}{45} ( \frac{1(1 - \frac{1}{10 {}^{n} } )}{(1 - \frac{1}{10} )} ) \\ becoz \: \: the \: \: above \: \: series \: is \: \: in \: \: g.p \\ \\ s = \frac{4n}{9} - \frac{2 \times 10}{45} ( \frac{10 {}^{n} - 1)}{10 {}^{n} \times 9} ) \\ \\ s = \frac{4n}{9} - \frac{20}{405} ( \frac{(10 {}^{n} - 1) }{10 {}^{n} } ) \\ \\ s = \frac{4n}{9} - \frac{4}{81} ( \frac{(10 {}^{n} - 1) }{10 {}^{n} } ) \\ \\ Therefore \: \: the \: \: sum \: \: of \: \: series \: \: \\ 0.4 + 0.44 + 0.444 + 0.4444 + ... + n \\ is \: \: \: \: \: \frac{4n}{9} - \frac{4}{81} ( \frac{(10 {}^{n} - 1) }{10 {}^{n} } )$