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Question Number 22491 by Tinkutara last updated on 19/Oct/17

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

{(1 - 2 x)}^{- 1 / 2} is

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

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

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

(4) \frac{(2 r)!}{2^{r} (r + 1)! (r - 1)!}

Answered by ajfour last updated on 19/Oct/17

c o e f f . o f x^{r} i n t h e e x p a n s i o n o f

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

= \frac{1}{2^{r}} \times \frac{2 n (2 n - 2) (2 n - 4) . . (2 n - 2 r + 2)}{r!}

c o e f f . o f x^{r} i n t h e e x p a n s i o n o f

{(1 - 2 x)}^{- 1 / 2} i s t h e n

= \frac{1}{2^{r}} \times \frac{{(- 1)}^{r} [1 \times 3 \times 5 \times 7 \times \dots \times (2 r - 1)]}{r!} {(- 2)}^{r}

= \frac{1 \times 3 \times 5 \times 7 \times \dots . \times (2 r - 1)}{r!}

= \frac{1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times \dots (2 r - 1) \times (2 r)}{{(2)}^{r} \times r! \times r!}

= \frac{(2 r)!}{(2^{r}) {(r!)}^{2}} . o p t i o n (2) .

Commented by Tinkutara last updated on 19/Oct/17

Thank you very much Sir!

Thank you very much Sir!