Q49 / 155 · WBJEE 2016 · Mathematics

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
    $\frac{1}{(1-a)(1-x)}$
  • Correct answer
    C
    $\frac{1}{(1-x)(1-a x)}$
  • D
    $\frac{1}{(1+a x)(1-a)}$

Solution

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

Hence, the correct answer is option: 'C'.

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