If $\left( {x + \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}}} \right)\left( {{x^4} + \frac{1}{{{x^4}}}} \right)\left( {{x^8} + \frac{1}{{{x^8}}}} \right)\left( {{x^{16}} + \frac{1}{{{x^{16}}}}} \right)?$

Question

TuteeHUB · Accepted Answer

$\frac{{\left( {{x^8} - \frac{1}{{{x^8}}}} \right)}}{{x - \frac{1}{x}}}$

TuteeHUB · Answer

$\frac{{\left( {{x^{32}} - \frac{1}{{{x^{32}}}}} \right)}}{{x - \frac{1}{x}}}$

TuteeHUB · Answer

$\frac{{\left( {{x^{64}} - \frac{1}{{{x^{64}}}}} \right)}}{{x - \frac{1}{x}}}$

TuteeHUB · Answer

$\frac{{\left( {{x^{16}} - \frac{1}{{{x^{16}}}}} \right)}}{{x - \frac{1}{x}}}$

1.	If \(\left( {x + \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}}} \right)\left( {{x^4} + \frac{1}{{{x^4}}}} \right)\left( {{x^8} + \frac{1}{{{x^8}}}} \right)\left( {{x^{16}} + \frac{1}{{{x^{16}}}}} \right)?\)
A.	\(\frac{{\left( {{x^{32}} - \frac{1}{{{x^{32}}}}} \right)}}{{x - \frac{1}{x}}}\)
B.	\(\frac{{\left( {{x^8} - \frac{1}{{{x^8}}}} \right)}}{{x - \frac{1}{x}}}\)
C.	\(\frac{{\left( {{x^{64}} - \frac{1}{{{x^{64}}}}} \right)}}{{x - \frac{1}{x}}}\)
D.	\(\frac{{\left( {{x^{16}} - \frac{1}{{{x^{16}}}}} \right)}}{{x - \frac{1}{x}}}\)
Answer» B. \(\frac{{\left( {{x^8} - \frac{1}{{{x^8}}}} \right)}}{{x - \frac{1}{x}}}\)

If \(\left( {x + \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}}} \right)\left( {{x^4} + \frac{1}{{{x^4}}}} \right)\left( {{x^8} + \frac{1}{{{x^8}}}} \right)\left( {{x^{16}} + \frac{1}{{{x^{16}}}}} \right)?\)

Discussion

No Comment Found

Related MCQs

Reply to Comment