Question
If $$x > 0,\frac{{{x^n}}}{{1 + x + {x^2} + ..... + {x^{2n}}}}\,{\text{is}}$$
A.
$$ \leqslant \frac{1}{{2n + 1}}$$
B.
$$ < \frac{2}{{2n + 1}}$$
C.
$$^3\frac{1}{{2n + 1}}$$
D.
$$ > \frac{2}{{2n + 1}}$$
Answer :
$$ \leqslant \frac{1}{{2n + 1}}$$
Solution :
$$\eqalign{
& x + \frac{1}{x} \geqslant 2,.....,{x^n} + \frac{1}{{{x^n}}} \geqslant 2 \cr
& {\text{on adding}} \cr
& \left( {x + \frac{1}{x}} \right) + \left( {{x^2} + \frac{1}{{{x^2}}}} \right) + ..... + \left( {{x^n} + \frac{1}{{{x^n}}}} \right) \geqslant 2n, \cr
& \left( {\frac{1}{{{x^n}}} + \frac{1}{{{x^{n - 1}}}} + ..... + \frac{1}{x}} \right) + 1 + \left( {x + {x^2} + ..... + {x^n}} \right) \geqslant 1 + 2n \cr
& \frac{{\left( {1 + x + ..... + {x^{n - 1}} + {x^n}} \right) + {x^{n + 1}} + {x^{n + 2}} + ..... + {x^{2n}}}}{{{x^n}}} \geqslant 1 + 2n \cr
& \frac{{{x^n}}}{{1 + x + ..... + {2^{2n}}}} \leqslant \frac{1}{{1 + 2n}} \cr} $$