In the expansion of (1+x)^(20) if the coefficient of x^r is twice the coefficient of x^(r−1) , what the value of the coefficient?


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Question Number 103826 by bobhans last updated on 17/Jul/20

$I n t h e e x p a n s i o n o f {(1 + x)}^{20} i f t h e$ $c o e f f i c i e n t o f x^{r} i s t w i c e t h e c o e f f i c i e n t$ $o f x^{r - 1}, w h a t t h e v a l u e o f t h e$ $c o e f f i c i e n t ?$
	Answered by bramlex last updated on 17/Jul/20

	${(1 + x)}^{20} = \underset{r = 0}{\sum^{20}} C_{r}^{20} 1^{r} . x^{20 - r}$ $c o e f f i c i e n t o f x^{r} i s (\begin{matrix} 20 \\ r \end{matrix}) a n d c o e f f i c i e n t o f x^{r - 1} i s (\begin{matrix} 20 \\ r - 1 \end{matrix})$ $s o c o n d i t i o n i n e q u a t i o n$ $(\begin{matrix} 20 \\ r \end{matrix}) = 2 (\begin{matrix} 20 \\ r - 1 \end{matrix})$ $n o t e (\begin{matrix} n \\ r \end{matrix}) = \frac{n - r + 1}{r} (\begin{matrix} n \\ r - 1 \end{matrix})$ $s o y o u r e q u a t i o n r e d u c e s t o$ $\frac{20 - r + 1}{r} = 2 \Rightarrow r = 7 . t h e r e f o r e$ $t h e v a l u e o f c o e f f i c i e n t \frac{20!}{7! . 13!}$ $= \frac{20.19.18.17.16.15.14}{7.6.5.4.3.2.1}$
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