The-coefficient-of-x-r-in-the-expansion-of-1-4x-1-2-is-

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

The coefficient of x^{r} in the expansion of

{(1 - 4 x)}^{- 1 / 2} 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 \dots \times (\frac{1}{2} - (r - 1))

C_{r}^{\frac{1}{2}} = \frac{1}{r!} \times \frac{1}{2} \times (\frac{1 - 2}{2}) \times (\frac{1 - 2 \times 2}{2}) \times \dots \times (\frac{1 - 2 (r - 1)}{2})

C_{r}^{\frac{1}{2}} = \frac{{(- 1)}^{r - 1} \times 1 \times 3 \times 5 \dots \times (2 r - 3)}{2^{r} r!}

C_{r}^{\frac{1}{2}} = \frac{{(- 1)}^{r - 1} (2 r - 3)!!}{2^{r} r!}

\sqrt{1 - x} = {(1 - x)}^{\frac{1}{2}} = \underset{r = 0}{\sum^{\infty}} C_{r}^{\frac{1}{2}} {(- x)}^{r}

\sqrt{1 - x} = - \underset{r = 0}{\sum^{\infty}} \frac{(2 r - 3)!!}{2^{r} r!} x^{r}

\sqrt{1 - 4 x} = - \underset{r = 0}{\sum^{\infty}} \frac{2^{r} (2 r - 3)!!}{r!} x^{r}

\frac{d \dots}{d x} o n b o t h s i d e s

\frac{1}{\sqrt{1 - 4 x}} = \underset{r = 1}{\sum^{\infty}} \frac{2^{r - 1} (2 r - 3)!!}{(r - 1)!} x^{r - 1}

{(1 - 4 x)}^{- \frac{1}{2}} = \frac{1}{\sqrt{1 - 4 x}} = \underset{r = 0}{\sum^{\infty}} \frac{2^{r} (2 r - 1)!!}{r!} x^{r}

\Rightarrow c o e f . o f x^{r} i s a_{r} = \frac{2^{r} (2 r - 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

\dots \dots

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