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-

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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