The coefficient of $x^{2^{m + 1}}$ in the expansion of $E = \frac{1}{(1 + x)\left( 1 + x^{2} \right)\left( 1 + x^{4} \right)\left( 1 + x^{8} \right)\ldots(1 + x^{2^{m}})}\ ,\ |x|

Question

The coefficient of $x^{2^{m + 1}}$ in the expansion of $E = \frac{1}{(1 + x)\left( 1 + x^{2} \right)\left( 1 + x^{4} \right)\left( 1 + x^{8} \right)\ldots(1 + x^{2^{m}})}\ ,\ |x| < 1$ is

Dr. Abubacker · Accepted Answer

(c)
Multiplying the numerator and denominator by $1 - x$, we have $E = \frac{1 - x}{(1 - x)(1 + x)\left( 1 + x^{2} ight)\left( 1 + x^{4} ight)\ldots(1 + x^{2^{m}})}$
$= \frac{1 - x}{\left( 1 - x^{2} ight)\left( 1 + x^{2} ight)\left( 1 + x^{4} ight)\ldots\left( 1 - x^{2^{m}} ight)}$
$= \frac{1 - x}{\left( 1 - x^{4} ight)\left( 1 + x^{4} ight)\ldots\left( 1 - x^{2^{m}} ight)}$
$= \frac{1 - x}{\left( 1 - x^{2^{m + 1}} ight)} = (1 - x){(1 - x^{2^{m + 1}})}^{- 1}$
$= (1 - x)(1 + x^{2^{m + 1}} + x^{2^{m + 2}} + ...)$
$	herefore$ Coefficient of $x^{2^{m + 1}}$ is 1

Q. The coefficient of $x^{2^{m + 1}}$ in the expansion of $E = \frac{1}{(1 + x)\left( 1 + x^{2} \right)\left( 1 + x^{4} \right)\left( 1 + x^{8} \right)\ldots(1 + x^{2^{m}})}\ ,\ |x| < 1$ is

Explanation

Reviewed by Dr. Abubacker

Want more practice?