If x-1-a-a2- - - to -a -1- y-1-b-b2- - - to -b -1- then Z-1-a b-a2 b2-a3 b3- - to - is

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Properties of Modulus

If x=1+a+a2……...

Question

If x = 1 + a + $a^{2}$ .......... to $\infty$ (|a|<1), y = 1 + b + $b^{2}$ ......... to $\infty$ (|b| < 1), then Z = 1 + ab + $a^{2}$ $b^{2}$ + $a^{3}$ $b^{3}$ ..... to ∞ is

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Solution

The correct option is C

We want to find Z in this problem. It is going to be in terms of 1 and b.

But the options are in terms of x and y.

We can find a and b in terms of x and y from the first two infinite series.

x = $\frac{1}{1 - a}$ -------------(1)

y = $\frac{1}{1 - b}$ --------------(2)

z = $\frac{1}{1 - a b}$ ----------(3)

From (1), 1 - a = $\frac{1}{x}$

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

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

Similarly b = $\frac{y - 1}{y}$

∴ z = $\frac{1}{1 - \frac{(x - 1)}{x} \frac{(y - 1)}{y}}$

= $\frac{x y}{x y - (x y - x - y + 1)}$

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

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