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Question Number 63937 by gunawan last updated on 11/Jul/19

$$\mathrm{The}\mathrm{coefficient}\mathrm{of}{x}^r\mathrm{in}\mathrm{the}\mathrm{expansion}\mathrm{of}$$$${(1-4x)}^{-1/2}\mathrm{is}$$

Answered by mr W last updated on 12/Jul/19

$$C_r^{\frac{1}{2}}=\frac{1}{r!}\times \frac{1}{2}\times (\frac{1}{2}-1)\times (\frac{1}{2}-2)\times ...\times (\frac{1}{2}-(r-1))$$$$C_r^{\frac{1}{2}}=\frac{1}{r!}\times \frac{1}{2}\times \left(\frac{1-2}{2}\right)\times \left(\frac{1-2\times 2}{2}\right)\times ...\times \left(\frac{1-2(r-1)}{2}\right)$$$$C_r^{\frac{1}{2}}=\frac{{(-1)}^{r-1}\times 1\times 3\times 5...\times (2r-3)}{2^{r}r!}$$$$C_r^{\frac{1}{2}}=\frac{{(-1)}^{r-1}(2r-3)!!}{2^{r}r!}$$$$\sqrt{1-x}={(1-x)}^{\frac{1}{2}}=\underset{r=0}{\sum ^{\mathrm{\infty }}}{C}_r^{\frac{1}{2}}{(-x)}^r$$$$\sqrt{1-x}=-\underset{r=0}{\sum ^{\mathrm{\infty }}}\frac{(2r-3)!!}{2^{r}r!}{x}^r$$$$\sqrt{1-4x}=-\underset{r=0}{\sum ^{\mathrm{\infty }}}\frac{2^r(2r-3)!!}{r!}{x}^r$$$$\frac{d...}{dx}onbothsides$$$$\frac{1}{\sqrt{1-4x}}=\underset{r=1}{\sum ^{\mathrm{\infty }}}\frac{2^{r-1}(2r-3)!!}{(r-1)!}{x}^{r-1}$$$${(1-4x)}^{-\frac{1}{2}}=\frac{1}{\sqrt{1-4x}}=\underset{r=0}{\sum ^{\mathrm{\infty }}}\frac{2^r(2r-1)!!}{r!}{x}^r$$$$\Rightarrow coef.of{x}^{r}is{a}_r=\frac{2^r(2r-1)!!}{r!}$$$$r=0:a_0=\frac{2^0\times 1}{1}=1$$$$r=1:a_1=\frac{2^1\times 1}{1}=2$$$$r=2:a_2=\frac{2^2\times 3\times 1}{2\times 1}=6$$$$r=3:a_3=\frac{2^3\times 5\times 3\times 1}{3\times 2\times 1}=20$$$$r=4:a_4=\frac{2^4\times 7\times 5\times 3\times 1}{4\times 3\times 2\times 1}=70$$$$r=5:a_5=\frac{2^5\times 9\times 7\times 5\times 3\times 1}{5\times 4\times 3\times 2\times 1}=252$$$$......$$

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