Sequence And Series Question 343
Question: If $ (1-p)(1+3x+9x^{2}+27x^{3}+81x^{4}+243x^{5}) $ $ =1-p^{6}, $ $ p\ne 1 $ then the value of $ \frac{p}{x} $ is
Options:
A) $ \frac{1}{3} $
B) 3
C) $ \frac{1}{2} $
D) 2
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ (1-p)(1+3x+9x^{2}+27x^{3}+81x^{4}+243x^{5}) $ $ =1-p^{6} $
$ \Rightarrow (1+3x+9x^{2}+27x^{3}+81x^{4}+243x^{5})=\frac{1-p^{6}}{1-p} $
$ \Rightarrow (1+3x+9x^{2}+27x^{3}+81x^{4}+243x^{5}) $ $ =1+p+p^{2}+p^{3}+p^{4}+p^{5} $ Comparing we get $ p=3x $ or $ p/x=3 $
BETA
