A) \[\cdot \le \frac{1}{2n+1}\]
B) \[<\frac{1}{2n+1}\]
C) \[^{3}\frac{1}{2n+1}\]
D) \[>\frac{2}{2n+1}\]
Correct Answer: A
Solution :
[a] \[x+\frac{1}{x}\ge 2,....,{{x}^{n}}+\frac{1}{{{x}^{n}}}\ge 2\] on adding \[\left( x+\frac{1}{x} \right)+\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)+....+\left( {{x}^{n}}+\frac{1}{{{x}^{n}}} \right)\ge 2n,\] \[\left( \frac{1}{{{x}^{n}}}+\frac{1}{{{x}^{n-1}}}+....\frac{1}{x} \right)+1+(x+{{x}^{2}}+....+{{x}^{n}})\ge 1+2n\]\[\frac{(1+x+...+{{x}^{n-1}}+{{x}^{n}})+{{x}^{n+1}}+{{x}^{n+2}}+...+{{x}^{2n}}}{{{x}^{n}}}\] \[\ge 1+2n\] \[\frac{{{x}^{n}}}{1+x+.....+{{2}^{2n}}}\le \frac{1}{1+2n}\]You need to login to perform this action.
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