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} = \fr

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

Similar Questions

$1+\sin x+\sin ^2 x+\sin ^3 x+\ldots+\infty=4+2 \sqrt{3}$ and $0 < x < \pi, x \neq \frac{\pi}{2}$,then $x=$

The sum of $n$ terms of the following series $1 \times 2 + 2 \times 3 + 3 \times 4 + 4 \times 5 + \dots$ is:

$\frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \frac{1}{8 \cdot 11} + \ldots$ to $16$ terms $=$

The $n^{th}$ term of the series $2 + 4 + 7 + 11 + \dots$ is

The largest perfect square that divides $2014^3 - 2013^3 + 2012^3 - 2011^3 + \ldots + 2^3 - 1^3$ is (in $^2$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module