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(C) The given series is $S = \sum_{n=0}^{\infty} (1 + a + a^2 + \dots + a^n)x^n$. We know that $(1 + a + a^2 + \dots + a^n) = \frac{1 - a^{n+1}}{1 - a

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$\frac{1}{(1 - x)(1 - ax)}$

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(C) The given series is $S = \sum_{n=0}^{\infty} (1 + a + a^2 + \dots + a^n)x^n$. We know that $(1 + a + a^2 + \dots + a^n) = \frac{1 - a^{n+1}}{1 - a}$. Substituting this into the sum: $S = \sum_{n=0}^{\infty} \frac{1 - a^{n+1}}{1 - a} x^n = \frac{1}{1 - a} \left[ \sum_{n=0}^{\infty} x^n - \sum_{n=0}^{\infty} a^{n+1} x^n \right]$. Using the sum of an infinite geometric series $\sum_{n=0}^{\infty} r^n = \frac{1}{1 - r}$ for $|r| $S = \frac{1}{1 - a} \left[ \frac{1}{1 - x} - a \sum_{n=0}^{\infty} (ax)^n \right]$. $S = \frac{1}{1 - a} \left[ \frac{1}{1 - x} - \frac{a}{1 - ax} \right]$. $S = \frac{1}{1 - a} \left[ \frac{1 - ax - a(1 - x)}{(1 - x)(1 - ax)} \right]$. $S = \frac{1}{1 - a} \left[ \frac{1 - ax - a + ax}{(1 - x)(1 - ax)} \right]$. $S = \frac{1}{1 - a} \left[ \frac{1 - a}{(1 - x)(1 - ax)} \right]$. $S = \frac{1}{(1 - x)(1 - ax)}$.

$1 + (1 + a)x + (1 + a + a^2)x^2 + \dots \infty = \dots \, (0 < a, x < 1)$

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