# NIELIT 2019 Feb Scientist D - Section D: 28

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$\left [\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^{2}}+\frac{4}{1+x^{4}}+\frac{8}{1+x^{8}} \right ]$ equal to :

1. $1$
2. $0$
3. $\frac{8}{1-x^{8}}$
4. $\frac{16}{1-x^{16}}$

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## 1 Answer

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$\left [ \frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}\right ]$

$\implies \left [ \frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}\right ]$

$\implies \left [ \frac{4}{1-x^4}+\frac{4}{1+x^4}+\frac{8}{1+x^8}\right ]$

$\implies \left [ \frac{8}{1-x^8}+\frac{8}{1+x^8}\right ]$

$\implies \left [ \frac{16}{1-x^{16}}\right ]$

Note: take the first 2 term's in each steps and solve them using LCM.

Option $D$ is correct here.
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