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Question

If x=1+a+a2+..., (|a|<1),y=1+b+b2+...,(|b|<1),  then z=1+ab+a2b2+a3b3+... is 
  1. x+y-1xy
  2. 2xyx+y-1
  3. xyx+y-1
  4. x+y-12xy


Solution

The correct option is C xyx+y-1

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=11a ... (1)

y=11b ... (2)

z=11ab ... (3)

From (1), 1a=1x

a=11x

 =x1x

Similarly, b=y1y

z=11(x1)x(y1)y

=xyxy(xyxy+1)

=xyx+y1

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