Sequence-And-Series Question 663
Question: If $ x>0,\frac{x^{n}}{1+x+x^{2}+…+x^{2n}} $ is
Options:
A) $ \cdot \le \frac{1}{2n+1} $
B) $ <\frac{1}{2n+1} $
C) $ ^{3}\frac{1}{2n+1} $
D) $ >\frac{2}{2n+1} $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ x+\frac{1}{x}\ge 2,….,x^{n}+\frac{1}{x^{n}}\ge 2 $ on adding $ ( x+\frac{1}{x} )+( x^{2}+\frac{1}{x^{2}} )+….+( x^{n}+\frac{1}{x^{n}} )\ge 2n, $ $ ( \frac{1}{x^{n}}+\frac{1}{{x^{n-1}}}+….\frac{1}{x} )+1+(x+x^{2}+….+x^{n})\ge 1+2n $ $ \frac{(1+x+…+{x^{n-1}}+x^{n})+{x^{n+1}}+{x^{n+2}}+…+x^{2n}}{x^{n}} $ $ \ge 1+2n $ $ \frac{x^{n}}{1+x+…..+2^{2n}}\le \frac{1}{1+2n} $
BETA
