If a, x are real numbers and |a|

Question

(C) The given series is $S = \sum_{n=1}^{\infty} (1+a+a^2+\dots+a^{n-1})x^{n-1}$.Using the sum of a geometric progression, $1+a+a^2+\dots+a^{n-1} = \f

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The given series is $S = \sum_{n=1}^{\infty} (1+a+a^2+\dots+a^{n-1})x^{n-1}$.Using the sum of a geometric progression, $1+a+a^2+\dots+a^{n-1} = \frac{1-a^n}{1-a}$.So, $S = \sum_{n=1}^{\infty} \frac{1-a^n}{1-a} x^{n-1} = \frac{1}{1-a} \sum_{n=1}^{\infty} (x^{n-1} - a^n x^{n-1})$.$S = \frac{1}{1-a} \left( \sum_{n=1}^{\infty} x^{n-1} - a \sum_{n=1}^{\infty} (ax)^{n-1} \right)$.Using the sum of an infinite geometric series $\sum_{k=0}^{\infty} r^k = \frac{1}{1-r}$ for $|r| $S = \frac{1}{1-a} \left( \frac{1}{1-x} - \frac{a}{1-ax} \right)$.$S = \frac{1}{1-a} \left( \frac{(1-ax) - a(1-x)}{(1-x)(1-ax)} \right) = \frac{1}{1-a} \left( \frac{1-ax-a+ax}{(1-x)(1-ax)} \right)$.$S = \frac{1}{1-a} \left( \frac{1-a}{(1-x)(1-ax)} \right) = \frac{1}{(1-x)(1-ax)}$.

If $a, x$ are real numbers and $|a| < 1, |x| < 1$, then $1 + (1+a)x + (1+a+a^2)x^2 + \dots \infty$ is equal to

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