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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Sequence and Series
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$p\;(x)=\large\frac{1+x^2+x^4+\;...\;x^{2n-2}}{1+x+x^2+\;...\;+x^{n-1}}\;$ is a polynomial in x , then n must be

$(a)\;even \qquad(b)\;even\qquad(c)\;odd\;than\;10\qquad(d)\;greater\;than\;10$

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1 Answer

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Answer : odd
Explanation : $p\;(x)=\large\frac{(1-x^{2n})}{(1-x^2)}$$\times\large\frac{(1-x)}{(1-x^n)}$
$=\large\frac{(1+x^{n})}{(1+x)}$
For $\;p\;(x)\;$ to be a polynomial , n must be odd.
answered Jan 22, 2014 by yamini.v
edited Mar 26, 2014 by balaji.thirumalai
 

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