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If x=1+y+y2……...

Question

If x = 1 + y + $y^{2}$ . . . . . $\infty$ , |y|<1, then y is

A

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C

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Solution

The correct option is C

x = $\frac{1}{1 - y}$ [ sum of an infinite G.P.]

$\Rightarrow$ 1 - y = $\frac{1}{x}$

$\Rightarrow$ y = 1 - $\frac{1}{x}$

= $\frac{x - 1}{x}$

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