If $| x | < 1$ , then the sum of the series $1 + 2 x + 3 x^{2} + 4 x^{3} + \dots$ , will be

\frac{1}{1 - x}

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\frac{1}{1 + x}

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\frac{1}{(1 + x)^{2}}

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\frac{1}{(1 - x)^{2}}

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Solution

The correct option is C $\frac{1}{(1 - x)^{2}}$
Let $S = 1 + 2 x + 3 x^{2} + 4 x^{3} + \dots$ ........(i)
$∴ x S = x + 2 x^{2} + 3 x^{3} + 4 x^{4} + \dots$ ........(ii)
Now subtracting (ii) from (i) by shifting one place, we get
$S (1 - x) = 1 + x + x^{2} + x^{3} + \dots = \frac{1}{1 - x}$
$∴ S = \frac{1}{(1 - x)^{2}}$
Hence, option 'D' is correct.

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