$1 + 2 x + 3 x^{2} + 4 x^{3} + . . . . . .$ = ?

\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}

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Solution

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

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Similar questions

Match the following (|x| < 1)

(1) ${(1 + x)}^{- 1}$ (P) 1 + 2x + 3 $x^{2}$ + 4 $x^{3}$ .........

(2) ${(1 - x)}^{- 1}$ (Q) 1 - x + $x^{2}$ - $x^{3}$ .........

(3) ${(1 + x)}^{- 2}$ (R) 1 + x + $x^{2}$ + $x^{3}$ .........

(4) ${(1 - x)}^{- 2}$ (S)1 - 2x + 3 $x^{2}$ - 4 $x^{3}$ .........

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

\int (x + x^{2} + x^{3} + x^{4})

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

3 x^{2}

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\infty

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