NIELIT 2019 Feb Scientist D - Section D: 28

Question

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

• A. $1$
• B. $0$
• C. $\frac{8}{1-x^{8}}$
• D. $\frac{16}{1-x^{16}}$

Answer 1

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

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

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

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

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

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

Option $D$ is correct here.