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

\frac{(3 r)!}{r! r!}

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\frac{(2 r)!}{r! r!}

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\frac{(4 r)!}{r! r!}

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None of the

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Solution

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

So, ${(1 - 4 x)}^{n} = \frac{n \cdot (n - 1) \cdot (n - 2) . . . \cdot (n - r + 1)}{r!} {(- 4)}^{r} x^{r}$
Now, for, $n = 1 / 2$ ,
${(1 - 4 x)}^{\frac{1}{2}} = \frac{\frac{1}{2} \cdot (\frac{1}{2} - 1) \cdot (\frac{1}{2} - 2) . . . . \cdot (\frac{1}{2} + 1 - r)}{r!} {(- 1)}^{r} x^{r} 4^{r}$
Hence the coefficient of $x^{r}$ is
$= \frac{\frac{1}{2} \cdot \frac{3}{2} \times . . . \times (\frac{2 r - 1}{2})}{r!} 4^{r}$
$= \frac{1 \times 3 \times 5 \times . . . . \times (2 r - 1)}{2^{r} r!} 4^{r} = \frac{(1 \times 3 \times 5 \times . . . . \times (2 r - 1)) \times 2 \times 4 \times 6 \times . . . \times 2 r}{2^{r} r! \times 2^{r} r!} 4^{r}$
$= \frac{(2 r)!}{r! r!}$
So, $B$ is correct.

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The coefficient of xr in the expansion of √1−4x.

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