Find-the-coefficient-of-x-11-in-2x-2-x-3-6-

Question Number 183205 by cortano1 last updated on 23/Dec/22

F i n d t h e c o e f f i c i e n t o f x^{11} i n

{(2 x^{2} + x - 3)}^{6} .

Answered by mr W last updated on 23/Dec/22

{(2 x^{2} + x - 3)}^{6} = \sum_{a + b + c = 6} (\begin{matrix} 6 \\ a, b, c \end{matrix}) {(2 x^{2})}^{a} x^{b} {(- 3)}^{c}

a + b + c = 6

2 a + b = 11

\Rightarrow b = 11 - 2 a ⩾ 0 \Rightarrow a ⩽ 5

\Rightarrow a = c + 5

\Rightarrow c = 0, a = 5, b = 1

c o e f . o f x^{11} :

2^{5} \times 1^{1} \times {(- 3)}^{0} \times \frac{6!}{5! \times 1! \times 0!} = 32 \times 6 = 192

Answered by cortano1 last updated on 24/Dec/22

\Rightarrow {(2 x^{2} + x - 3)}^{6} = 64 x^{12} {(1 + \frac{1}{2} x^{- 1} - \frac{3}{2} x^{- 2})}^{6}

= 64 x^{12} . \underset{0 ⩽ a, b, c ⩽ 6}{\sum^{a + b + c = 6}} (\begin{matrix} 6 \\ a, b, c \end{matrix}) (1^{a}) {(\frac{1}{2} x^{- 1})}^{b} {(- \frac{3}{2} x^{- 2})}^{c}

[- b - 2 c = - 1, b + 2 c = 1 \land a + b + c = 6]

∵ c = 0, b = 1, a = 5

C o e f f i c i e n t o f x^{11} i s 64 \times \frac{6!}{5! . 1! . 0!} \times \frac{1}{2}

= 64 \times 3 = 192

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