The-coefficient-of-x-4-in-the-expansion-of-1-x-x-2-x-3-11-is-

Question Number 55785 by gunawan last updated on 04/Mar/19

The coefficient of x^{4} in the expansion of

{(1 + x + x^{2} + x^{3})}^{11} is

Commented by Tawa1 last updated on 04/Mar/19

coefficient of x^{4} in {(1 + x + x^{2} + x^{3})}^{11} is \dots

= \frac{11!}{(α_{1})! (α_{2})! (α_{3})! (α_{4})!} 1^{α_{1}} {(x)}^{α_{2}} {(x^{2})}^{α_{3}} {(x^{3})}^{α_{4}}

= \frac{11!}{(α_{1})! (α_{2})! (α_{3})! (α_{4})!} {(x)}^{α_{2}} {(x)}^{2 α_{3}} {(x)}^{3 α_{4}}

= \frac{11!}{(α_{1})! (α_{2})! (α_{3})! (α_{4})!} {(x)}^{α_{2} + 2 α_{3} + 3 α_{4}}

where α_{1}, α_{2}, α_{3}, α_{4} are non - negative integer satisfy

α_{1} + α_{2} + α_{3} + α_{4} = 11 \dots \dots equation (i)

α_{2} + 2 α_{3} + 3 α_{4} = 4 \dots \dots equation (ii)

Hence, α_{1} = 9, α_{2} = 1, α_{3} = 0, α_{4} = 1

α_{1} = 9, α_{2} = 0, α_{3} = 2, α_{4} = 0

α_{1} = 8, α_{2} = 2, α_{3} = 1, α_{4} = 0

α_{1} = 7, α_{2} = 4, α_{3} = 0, α_{4} = 0

coefficient of x^{4} in {(1 + x + x^{2} + x^{3})}^{11} is now

= \frac{11!}{9! 1! 0! 1!} + \frac{11!}{9! 0! 2! 0!} + \frac{11!}{8! 2! 1! 0!} + \frac{11!}{7! 4! 0! 0!}

= \frac{11 \times 10 \times 9!}{9! 1! 0! 1!} + \frac{11 \times 10 \times 9!}{9! 0! 2! 0!} + \frac{11 \times 10 \times 9 \times 8!}{8! 2! 1! 0!} + \frac{11 \times 10 \times 9 \times 8 \times 7!}{7! 4! 0! 0!}

= 110 + \frac{110}{2!} + \frac{990}{2!} + \frac{7920}{4!}

= 110 + \frac{110}{2} + \frac{990}{2} + \frac{7920}{24}

= 110 + 55 + 495 + 330

= 990

Answered by tanmay.chaudhury50@gmail.com last updated on 04/Mar/19

{(1 + x) + x^{2} (1 + x)}^{11}

{(1 + x) (1 + x^{2})}^{11}

{(1 + x)}^{11} {(1 + x^{2})}^{11}

(1 + 11 c_{1} x + 11 c_{2} x^{2} + 11 c_{3} x^{3} + 11 c_{4} x^{4} + \dots + 11 c_{11} x^{11}) \times (1 + 11 c_{1} x^{2} + 11 c_{2} x^{4} + 11 c_{3} x^{6} + \dots + 11 c_{11} x^{22})

t e r m s c o n t a i n i n g x^{4} a r e

1 \times 11 c_{2} x^{4} + 11 c_{2} x^{2} \times 11 c_{1} x^{2} + 11 c_{4} x^{4} \times 1

= x^{4} (11 c_{2} + 11 c_{2} \times 11 c_{1} + 11 c_{4})

= x^{4} (\frac{11!}{2! 9!} + \frac{11!}{2! 9!} \times \frac{11!}{1! 10!} + \frac{11!}{4! 7!})

= x^{4} (55 + 55 \times 11 + \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2})

= x^{4} (55 + 605 + 330)

= x^{4} (990)

p l s c h e c k \dots

Answered by mr W last updated on 06/Mar/19

1 + x + x^{2} + x^{3} = \frac{1 - x^{4}}{1 - x}

{(1 + x + x^{2} + x^{3})}^{11}

= {(1 - x^{4})}^{11} {(1 - x)}^{- 11}

= \underset{k = 0}{\sum^{11}} C_{k}^{11} {(- x^{4})}^{k} \underset{k = 0}{\sum^{\infty}} C_{k}^{10 + k} x^{k}

c o e f . o f x^{4} :

C_{1}^{11} {(- 1)}^{1} + C_{4}^{14}

= - 11 + 1001

= 990

Commented by tanmay.chaudhury50@gmail.com last updated on 05/Mar/19

s u m o f g p s e r i e s \dots S = \frac{a (1 - r^{n})}{1 - r} = \frac{1 - x^{4}}{1 - x}

Commented by Tawa1 last updated on 05/Mar/19

Sir, please can you explain this method very well sir,

i like it sir . How 1 + x + x^{2} + x^{3} = \frac{1 - x^{4}}{1 - x} and the summation part . .

etc \dots

Commented by Tawa1 last updated on 05/Mar/19

God bless you sir . i appreciate

Commented by Tawa1 last updated on 05/Mar/19

Why is the summation zero to infinity for power - 11

Commented by mr W last updated on 06/Mar/19

\frac{1}{1 - x} = 1 + x + x^{2} + x^{3} + \dots \to i n f i n i t e

\frac{1}{{(1 - x)}^{p}} = {(1 + x + x^{2} + x^{3} + \dots)}^{p} \to i n f i n i t e

\frac{1}{{(1 - x)}^{p}} = \underset{k = 0}{\sum^{\infty}} a_{k} x^{k}

Commented by Tawa1 last updated on 06/Mar/19

God bless you sir

Commented by mr W last updated on 06/Mar/19

{i t}^{'} s u s e f u l :

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

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