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
- A \(\frac{1}{(1-a)(1-a x)}\)
- B \(\frac{1}{(1-a)(1-x)}\)
- C \(\frac{1}{(1-x)(1-a x)}\)
- D \(\frac{1}{(1+a x)(1-a)}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{(1-x)(1-a x)}\)
Step-by-step Solution
Detailed explanation
We have. \(1+(1+a) x+\left(1+a+a^{2}\right) x^{2}+\ldots \infty\) \(=\sum_{n=1}^{\infty}\left(1+a+a^{2}+\ldots+a^{n-1}\right) x^{n-1}\) \(=\sum_{n=1}^{\infty}\left(\frac{1-a^{n}}{1-a}\right) x^{n-1}\)…
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