If |x| < 1, then the sum of the series 1 + 2x + $3 x^{2}$ + $4 x^{3}$ + ....... $\infty$ 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 D $\frac{1}{(1 - x)^{2}}$
This is an A.G.P.
Let S = 1 + 2x + $3 x^{2} + . . . . \infty$
$\Rightarrow x . S = x + 2 x^{2} + . . . . . \infty$
Subtracting $(1 - x) S = 1 + x + x^{2} + . . . . . . \infty = \frac{1}{1 - x}$
$∴ S = \frac{1}{(1 - x)^{2}} .$
Aliter : Use S = $[1 + \frac{r}{1 - r} \times d i f f . o f A . P .] \frac{1}{1 - r} .$

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Similar questions

| x | < 1

1 + 2 x + 3 x^{2} + 4 x^{3} + . . . . . . . . . \infty

| x | < 1

1 + 2 x + 3 x^{2} + 4 x^{3} + \dots

x < 1

1 + 2 x + 3 x^{2} + 4 x^{3} + . . . .

If |x|<1, then the sum of the series

1 + 2x + 3 $x^{2}$ + 4 $x^{3}$ + ........., ∞ will be

3 x^{2}

4 x^{3}

\infty

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