Question

Find the sum of first $n$ terms of the series
(i) $7+77+777+....$
(ii) $0.4+0.94+0.994+.....$

Answer 1

(i) $7+77+777+...$

$=\frac{7}{9}(9+99+\mathrm{999...})=\frac{7}{9}(10-1+{10}^2-1+{10}^3-\mathrm{1...})=\frac{7}{9}(10\left(\frac{{10}^n-1}{9}\right)-n)$

(ii) $0.4+0.94+0.994+...$

$=1-{6.10}^{-1}+1-{6.10}^{-2}+1-{6.10}^{-3}...=n-0.6\left(\frac{1-{10}^{-n}}{0.9}\right)=n-\frac{2}{3}(1-{10}^{-n})$