If x = sum_{n = 0}^infty a^n,y = sum_{n = 0}^ | vedclass

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(B) Given $x = \sum_{n = 0}^\infty a^n = \frac{1}{1 - a}$.Rearranging for $a$,we get $1 - a = \frac{1}{x}$,so $a = 1 - \frac{1}{x} = \frac{x - 1}{x}$.

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$xz + yz = xy + z$

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(B) Given $x = \sum_{n = 0}^\infty a^n = \frac{1}{1 - a}$.Rearranging for $a$,we get $1 - a = \frac{1}{x}$,so $a = 1 - \frac{1}{x} = \frac{x - 1}{x}$.Similarly,$y = \sum_{n = 0}^\infty b^n = \frac{1}{1 - b}$,which gives $b = \frac{y - 1}{y}$.Also,$z = \sum_{n = 0}^\infty (ab)^n = \frac{1}{1 - ab}$,which gives $ab = \frac{z - 1}{z}$.Substituting the expressions for $a$ and $b$ into the expression for $ab$:$\left(\frac{x - 1}{x}\right) \left(\frac{y - 1}{y}\right) = \frac{z - 1}{z}$.$\frac{(x - 1)(y - 1)}{xy} = \frac{z - 1}{z}$.$z(xy - x - y + 1) = xy(z - 1)$.$xyz - xz - yz + z = xyz - xy$.$-xz - yz + z = -xy$.$xy + z = xz + yz$.

If $x = \sum_{n = 0}^\infty a^n$,$y = \sum_{n = 0}^\infty b^n$,and $z = \sum_{n = 0}^\infty (ab)^n$,where $a, b < 1$,then:

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