$a b + 2 a b^{2} + 3 a b^{3} + 4 a b^{4} + . . . . . .$ = ?

\frac{1}{1 - a b}

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\frac{1}{(1 - a b)^{2}}

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\frac{1}{1 - a b^{3}}

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\frac{a b}{(1 - b)^{2}}

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Solution

The correct option is D $\frac{a b}{(1 - b)^{2}}$
Let $S$ be the sum of given series,
$∴ S = a b + 2 a b^{2} + 3 a b^{3} + 4 a b^{4} + . . . . . . . . . .$

multiplying $b$ both sides, we get

$b S = a b^{2} + 2 a b^{3} + 3 a b^{4} + . . . . . . . . .$
subtracting,
$S (1 - b) = a b + a b^{2} + a b^{3} + a b^{4} + . . . . . . . . . . . . . .$
$S (1 - b) = \frac{a b}{1 - b}$ applying sum of a GP

$S = \frac{a b}{(1 - b)^{2}}$

Suggest Corrections

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