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Let x = 1 + a + a² + a³ + ⋯ and y = 1 + b + b² + b³ + ⋯, where a and b are proper fractions. Then the value of 1 + ab + a²b² + a³b³ + ⋯ is:

1. xy/(x+y-1)
2. xy/(x-y-1)
3. xy/(x-y+1)
4. xy/(x+y+1)

Correct answer: xy/(x+y-1)

Solution

x=1/(1-a), y=1/(1-b), and the target sum is 1/(1-ab). With a=(x-1)/x, b=(y-1)/y, 1-ab=(x+y-1)/(xy), so 1/(1-ab)=xy/(x+y-1).

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