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WBJEE · Maths · Sequences and Series

If \(a, x\) are real numbers and \(|a| < 1,|x| < 1\) then \(1 (1+a) x+\left(1+a+a^{2}\right) x^{2}+\ldots \infty \quad\) is equal to

  1. A \(\frac{1}{(1-a)(1-a x)}\)
  2. B \(\frac{1}{(1-a)(1-x)}\)
  3. C \(\frac{1}{(1-x)(1-a x)}\)
  4. D \(\frac{1}{(1+a x)(1-a)}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{1}{(1-x)(1-a x)}\)

Step-by-step Solution

Detailed explanation

We have. \(1+(1+a) x+\left(1+a+a^{2}\right) x^{2}+\ldots \infty\) \(=\sum_{n=1}^{\infty}\left(1+a+a^{2}+\ldots+a^{n-1}\right) x^{n-1}\) \(=\sum_{n=1}^{\infty}\left(\frac{1-a^{n}}{1-a}\right) x^{n-1}\)…