Find the sum to 'n' terms of 0.9+0.99 +0.999 +...........

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Solution

This can be written as
$(1 - 0.1) + (1 - 0.01) + (1 - 0.001) + . . . . . . . . n t e r m s = (1 + 1 + 1 + . . . . . . . n t e r m s) - (0.1 + 0.01 + 0.001 + . . . . . . . n t e r m s) = n + \frac{0.1 (1 - {0.1}^{n})}{1 - 0.1} = n + \frac{0.1 (1 - {0.1}^{n})}{0.9} = n + \frac{1 - {0.1}^{n}}{9}$

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