Question

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

Ans: (C)
 \(: \frac{1}{1-x}+\frac{a x}{1-x}+\frac{a^{2} x^{2}}{1-x}+\ldots .=\frac{1}{1-x} \times\left(1+a x+a^{2} x^{2}+\ldots\right)=\frac{1}{1-x} \cdot \frac{1}{1-a x}\)

Answer 2

buy tadalafil 20mg generic <a href="https://ordergnonline.com/">coupon for cialis</a> the best ed pill

Answer 3

buy cialis 5mg <a href="https://ordergnonline.com/">cheap tadalafil 10mg</a> medications for ed