Given x = 1 + a + a^2 + ... infty (a

Question

(A) The given series are infinite geometric progressions $(G.P.)$.For $x = 1 + a + a^2 + ... \infty$,the sum is $x = \frac{1}{1 - a}$.Rearranging for

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

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(A) The given series are infinite geometric progressions $(G.P.)$.For $x = 1 + a + a^2 + ... \infty$,the sum is $x = \frac{1}{1 - a}$.Rearranging for $a$,we get $1 - a = \frac{1}{x}$,so $a = 1 - \frac{1}{x} = \frac{x - 1}{x}$.Similarly,for $y = 1 + b + b^2 + ... \infty$,the sum is $y = \frac{1}{1 - b}$.Rearranging for $b$,we get $1 - b = \frac{1}{y}$,so $b = 1 - \frac{1}{y} = \frac{y - 1}{y}$.The required series is $1 + ab + a^2b^2 + ... \infty$,which is also an infinite $G.P.$ with first term $1$ and common ratio $ab$.The sum is $\frac{1}{1 - ab}$.Substituting the values of $a$ and $b$:Sum $= \frac{1}{1 - (\frac{x - 1}{x})(\frac{y - 1}{y})} = \frac{1}{1 - \frac{(x - 1)(y - 1)}{xy}}$.Sum $= \frac{xy}{xy - (xy - x - y + 1)} = \frac{xy}{xy - xy + x + y - 1} = \frac{xy}{x + y - 1}$.

Given $x = 1 + a + a^2 + ... \infty$ $(a < 1)$ and $y = 1 + b + b^2 + ... \infty$ $(b < 1)$,find the value of $1 + ab + a^2b^2 + ... \infty$.

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