What-is-the-coefficient-of-x-2020-in-1-x-x-2-x-3-x-2020-2021-

Question Number 163576 by cortano1 last updated on 08/Jan/22

W h a t i s t h e c o e f f i c i e n t o f x^{2020}

i n {(1 + x + x^{2} + x^{3} + \dots + x^{2020})}^{2021}

Answered by bobhans last updated on 08/Jan/22

f (x) = {(1 + x + x^{2} + \dots + x^{2020})}^{2021}

f (x) = {(\frac{1}{1 - x})}^{2021} = \underset{m = 0}{\sum^{2021}} (\begin{matrix} 2021 + m - 1 \\ m \end{matrix}) x^{k}

take k = 2020 \Rightarrow coefficient of x^{2020} = (\begin{matrix} 4040 \\ 2020 \end{matrix})

Answered by mr W last updated on 08/Jan/22

s i n c e y o u a s k f o r t h e c o e f . o f x^{2020},

y o u c a n a d d x^{2021}, x^{2022}, \dots, \infty i n t o t h e

e x p r e s s i o n . t h a t w o n t c h a n g e t h e

c o e f . o f x^{2020} .

t h a t m e a n s c o e f . o f x^{2020} i n

{(1 + x + x^{2} + x^{3} + \dots + x^{2020})}^{2021} i s t h e

s a m e a s c o e f . o f x^{2020} i n

{(1 + x + x^{2} + x^{3} + \dots + x^{2020} + x^{2021} + \dots)}^{2021} .

{(1 + x + x^{2} + x^{3} + \dots + x^{2020} + x^{2021} + \dots)}^{2021}

= {(\frac{1}{1 - x})}^{2021} = \underset{k = 0}{\sum^{\infty}} C_{2020}^{k + 2020} x^{k}

t h e c o e f . o f x^{2020} i s t h e n

C_{2020}^{2020 + 2020} = C_{2020}^{4040}

Commented by mr W last updated on 08/Jan/22

b u t i f y o u a s k f o r t h e c o e f . o f x^{2050},

t h e n y o u m u s t p r o c e e d n o r m a l l y

a s f o l l o w i n g .

{(1 + x + x^{2} + x^{3} + \dots + x^{2020})}^{2021}

= {[\frac{1 - x^{2021}}{1 - x}]}^{2021} = {(1 - x^{2021})}^{2021} \underset{k = 0}{\sum^{\infty}} C_{k}^{k + 2020} x^{k}

c o e f . o f x^{2020} i s C_{2020}^{2020 + 2020}

c o e f . o f x^{2050} i s C_{2050}^{2050 + 2020} - 2021 C_{29}^{29 + 2020}

Commented by Tawa11 last updated on 08/Jan/22

Great sir

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