Question

Find the sum to $n$ terms $0.7+0.77+0.777+....$

Answer 1

To find: $0.7+0.77+0.777+.....$

We know that,

$0.1+{0.1}^2+{0.1}^3+...=\frac{0.1\times (1-{0.1}^n)}{0.9}=\frac{(1-{0.1}^n)}{9}=7\times (0.1+0.11+0.111+..)=7\times (\frac{1}{10}+\frac{11}{100}+\frac{111}{1000}+...)=\frac{7}{9}\times (\frac{9}{10}+\frac{99}{100}+\frac{999}{1000}+..)=\frac{7}{9}\times \left(\right(1-\frac{1}{10})+(1-\frac{1}{100})+(1-\frac{1}{1000})+..)=\frac{7}{9}\times \left(\right(1+1+...\text{n terms})-(\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+...\text{n terms}))=\frac{7}{9}\times (n-\frac{(1-{0.1}^n)}{9})$