Question

Coefficient of $x^{2^{m+1}}$ in the expansion of
$E=$ $\frac{1}{(1+x)(1+x^2)(1+x^4)(1+x^8)....(1+x^{2m})}(\left|x\right|<1)$

• A. $3$
• B. $2$
• C. $1$
• D. $0$

Answer 1

Answer: (C)

$1$

Multiplying and dividing the above expression by $(1-x)$
Therefore
$\frac{1-x}{(1-x)(1+x)(1+x^2)(1+x^4)...(1+x^{2^m})}$
$=\frac{1-x}{(1-x^2)(1+x^2)(1+x^4)...(1+x^{2^m})}$
$=\frac{1-x}{(1-x^4)(1+x^4)...(1+x^{2^m})}$
:
:
:
$=\frac{1-x}{(1-x^{2^m})(1+x^{2^m})}$
$=\frac{1-x}{(1-x^{2^{m+1}})}$
$=(1-x)(1-x^{2^{m+1}}{)}^{-1}$
$=(1-x)(1+x^{2^{m+1}}+(x^{2^{m+1}}{)}^2...\infty )$ ...(since $|x|<1$)
$=(1+x^{2^{m+1}}+(x^{2^{m+1}}{)}^2...\infty -(x+x^{2^{m+1}+1}...\infty )$
Hence coefficient of $x^{2^{m+1}}$ is $1$.