Question

The coefficient of $x^{49}$ in the expansion of $(x-1)(x-\frac{1}{2})(x-\frac{1}{2^2})..........(x-\frac{1}{2^{49}})$ is equal to.

• A. $-2(1-\frac{1}{2^{50}})$
• B. $+vecoefficientofx$.
• C. $-vecoefficientofx$.
• D. $-2(1-\frac{1}{2^{49}})$

Answer 1

The correct option is A $-2(1-\frac{1}{2^{50}})$

Consider $(x-1)(x-\frac{1}{2})(x-\frac{1}{2^2})...(x-\frac{1}{2^{49}})$

Coefficient of $x^{49}=-1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{49}}$

$=-[1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}]$

$=-\left[\frac{1({\left(\frac{1}{2}\right)}^{50}-1)}{\frac{1}{2}-1}\right]$

$=-\left[\frac{2^{50}-1}{2^{50}\left(\frac{2-1}{2}\right)}\right]$

$=-\left[\frac{2^{50}-1}{2^{50}\left(\frac{1}{2}\right)}\right]$

$=-\left[\frac{2^{50}-1}{2^{49}}\right]$

$=-2[1-\frac{1}{2^{50}}]$