$x + 2 x^{2} + 3 x^{3} + 4 x^{4} + . . . . . .$ = ?

\frac{1}{1 - x}

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

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\frac{1}{1 - x^{3}}

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

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Solution

The correct option is D $\frac{x}{(1 - x)^{2}}$
Let $S = x + 2 x^{2} + 3 x^{3} + 4 x^{4} + . . . . . . . . . .$
Multiplying by $x$ both sides,
$x S = x^{2} + 2 x^{3} + 3 x^{4} + . . . . . . . . .$
Subtracting both the equations:
$S (1 - x) = x + x^{2} + x^{3} + x^{4} + . . . . . . . . . . . . .$
$S (1 - x) = \frac{x}{1 - x}$ applying sum of GP for $\infty$ terms
$S = \frac{x}{(1 - x)^{2}}$

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