Question

The coefficient $x^{49}$ 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. $+$ve coefficient of $x$
• C. $-$ve coefficient of $x$
• D. $-2(1-\frac{1}{2^{49}})$

Answer 1

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

If $a,ar,a{r}^2,a{r}^3,.....,a{r}^{n-1}$ are n terms following geometric progression with common ratio $r$, $a+ar+a{r}^2+a{r}^3+...\dots \dots ..+a{r}^{n-1}=\frac{a(1-r^n)}{1-r}$ where $|r|<1$.

coefficient of $x^{49}$ in$(x-1)(x-\frac{1}{2})(x-\frac{1}{2^2})\dots \dots \dots ..(x-\frac{1}{2^{49}})$ is $-(1+\frac{1}{2}+\frac{1}{2^2}+...\dots \dots \dots \dots .+\frac{1}{2^{49}})=\frac{-(1-\frac{1}{2^{50}})}{1-\frac{1}{2}}=-2(1-\frac{1}{2^{50}})$