Sequence And Series Question 46
Question: The sum of $ n $ terms of the following series $ 1+(1+x)+(1+x+x^{2})+………. $ will be
[IIT 1962]
Options:
A) $ \frac{1-x^{n}}{1-x} $
B) $ \frac{x(1-x^{n})}{1-x} $
C) $ \frac{n(1-x)-x(1-x^{n})}{{{(1-x)}^{2}}} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ 1+(1+x)+(1+x+x^{2})+… $ + $ (1+x+x^{2}+x^{3}+…+{x^{n-1}})+… $ Required sum = $ \frac{1}{(1-x)}{ ,(1-x)+(1-x^{2})+(1-x^{3}) . $ $ . +(1-x^{4})+……….uptp\ n\ terms } $ $ =\frac{1}{(1-x)}[n-{x+x^{2}+x^{3}+……….upto\ n\ \text{terms }},] $ $ =\frac{1}{(1-x)}[ n-\frac{x(1-x^{n})}{1-x} ]=\frac{n(1-x)-x(1-x^{n})}{{{(1-x)}^{2}}} $ .
BETA
