The number of natural number n in the interva | vedclass

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Question

(c)

Let $P(x)=1+x^2+x^4+\ldots+x^{2010}$

${c}P(x)=\frac{1-x^{2012}}{1-x^2}$

$P(x)=\frac{\left(1-x^{1006}\right)\left(1+x^{1006}\right)}{(1-x)

Vedclass · Accepted Answer

$503$

Vedclass · Answer

(c)

Let $P(x)=1+x^2+x^4+\ldots+x^{2010}$

${c}P(x)=\frac{1-x^{2012}}{1-x^2}$

$P(x)=\frac{\left(1-x^{1006}\right)\left(1+x^{1006}\right)}{(1-x)(1+x)}$$P(x)=\left(1+x^{1006}\right)\left(\frac{1-x^{503}}{1-x}\right)\left(\frac{1+x^{503}}{1+x}\right)$

$\left.P(x)=\left(1+x^{1006}\right) 1+x+x^2+x^3+\ldots+x^{502}\right)$

$\left(1-x+x^2-x^3+\ldots+x^{502}\right)$

Thus, $P(x)$ is divisible by $1+x+x^2+\ldots x^{n-1}$

If $n-1=502 \Rightarrow n=503$

The number of natural number $n$ in the interval $[1005, 2010]$ for which the polynomial. $1+x+x^2+x^3+\ldots+x^{n-1}$ divides the polynomial $1+x^2+x^4+x^6+\ldots+x^{2010}$ is

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