# The sum of n terms of the series 4 + 44 + 444 + ... is

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GATE EC 2011 Official Paper
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1. (4/81) [10n+1 - 9n - 1]
2. (4/81) [10n-1 - 9n - 1]
3. (4/81) [10n+1 - 9n - 10]
4. (4/81) [10n - 9n - 10]

Let S = 4(1 + 11 + 111 + ...) = $$\frac{4}{9}(9 + 99 + 999 + ...)$$
$$\Rightarrow\frac{4}{9}[(10 - 1) + (10^2-1)+(10^3-1)+...]$$
$$\Rightarrow\frac{4}{9}[(10 + 10^2 +...10^n)-n]=\frac{4}{9}[10\frac{(10^n-1)}{9}]$$
$$\Rightarrow \frac{4}{81}[10^{n+1}-9n-10]$$
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