Question

The coefficient of $x^r$ in the expansion of $\frac{1}{\sqrt{1-4x}}$.

• A. $\frac{\left(r\right)!}{r!r!}$
• B. $\frac{\left(2r\right)!}{r!r!}$
• C. $\frac{\left(2r\right)!}{r!}$
• D. None of the above

Answer 1

The correct option is B $\frac{\left(2r\right)!}{r!r!}$

We know, ${(1+x)}^n=\frac{n\cdot (n-1)\cdot (n-2)...\cdot (n-r+1)}{r!}{x}^r$

So, ${(1-4x)}^n=\frac{n\cdot (n-1)\cdot (n-2)...\cdot (n-r+1)}{r!}{(-4)}^r{x}^r$

Now, for, $n=-1/2$ ,

${(1-4x)}^{\frac{-1}{2}}=\frac{\frac{-1}{2}\cdot (\frac{-1}{2}-1)\cdot (\frac{-1}{2}-2)....\cdot (\frac{-1}{2}+1-r)}{r!}{(-1)}^r{x}^r{4}^r$

Hence the coefficient of $x^r$ is

$=\frac{\frac{1}{2}\cdot \frac{3}{2}\times ...\times \left(\frac{2r-1}{2}\right)}{r!}{4}^r$

$=\frac{1\times 3\times 5\times ....\times (2r-1)}{2^{r}r!}{4}^r=\frac{(1\times 3\times 5\times ....\times (2r-1))\times 2\times 4\times 6\times ...\times 2r}{2^{r}r!\times 2^{r}r!}{4}^r$

$=\frac{\left(2r\right)!}{r!r!}$

So, $B$ is correct.