Sequence And Series Question 471
Question: If $ | x |<\frac{1}{2}, $ what is the value of $ 1+n[ \frac{x}{1-x} ]+[ \frac{n(n+1)}{2!} ]{{[ \frac{x}{1-x} ]}^{2}}+…….\infty $ ?
Options:
A) $ {{[ \frac{1-x}{1-2x} ]}^{n}} $
B) $ {{(1-x)}^{n}} $
C) $ {{[ \frac{1-2x}{1-x} ]}^{n}} $
D) $ {{( \frac{1}{1-x} )}^{n}} $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Given that $ 1+n| \frac{x}{1-x} |+\frac{n(n+1)}{2!}{{| \frac{x}{1-x} |}^{2}} $ $ +….\infty $ is expansion of $ {{| 1-\frac{x}{1-x} |}^{-n}} $ . So, it is $ ={{| 1-\frac{x}{1-x} |}^{-n}} $ $ ={{| \frac{1-x-x}{1-x} |}^{-n}}={{| \frac{1-x}{1-2x} |}^{n}} $
BETA
