Let $${S_n}\left( {1 \leqslant n \leqslant 9} \right)$$   denotes the sum of $$n$$ terms of series $$1 + 22 + 333 + . . . . . + 9999999999,$$       then for $${2 \leqslant n \leqslant 9}$$

Solution :
$$\eqalign{ & {S_n} = \frac{1}{9}\left( 9 \right) + \frac{2}{9}\left( {99} \right) + \frac{3}{9}\left( {999} \right) + .... \cr & = \frac{1}{9}\left[ {10 + {{2.10}^2} + {{3.10}^3} + ....} \right] - \frac{1}{9}\left[ {1 + 2 + 3 + ....} \right] \cr & = \frac{1}{9}S - \frac{1}{9}\frac{{n\left( {n + 1} \right)}}{2} \cr & S = 10 + {2.10^2} + {3.10^3} + .... + n{.10^n} \cr & \frac{{10S = {{10}^2} + {{2.10}^3} + .... + \left( {n - 1} \right){{10}^n} + n{{.10}^{n + 1}}}}{{ - 9S = \left( {10 + {{10}^2} + .... + {{10}^n}} \right) - n{{.10}^{n + 1}}}} \cr & S = \frac{n}{9}{10^{n + 1}} - \frac{{{{10}^{n + 1}} - 10}}{{81}} \cr & \therefore {S_n} = \frac{n}{{81}}{10^{n + 1}} - \frac{{{{10}^{n + 1}} - 10}}{{9.81}} - \frac{1}{9}\frac{{n\left( {n + 1} \right)}}{2} \cr & \therefore 9{S_n} = \frac{{\left( {9n - 1} \right){{10}^{n + 1}}}}{{81}} + \frac{{10}}{{81}} - \frac{{n\left( {n + 1} \right)}}{2} \cr & \therefore 9\left( {{S_n} - {S_{n - 1}}} \right) = \frac{{{{10}^n}}}{{81}}\left\{ {10\left( {9n - 1} \right) - \left( {9n - 10} \right)} \right\} - n = n\left( {{{10}^n} - 1} \right) \cr} $$