Coefficient-of-x-5-in-the-expansion-of-x-2-x-2-5-is-

Question Number 32541 by rahul 19 last updated on 27/Mar/18

C o e f f i c i e n t o f x^{5} i n t h e e x p a n s i o n

o f {(x^{2} - x - 2)}^{5} i s

Answered by MJS last updated on 27/Mar/18

= {(x - 2)}^{5} {(x + 1)}^{5}

coefficients of {(a \pm b)}^{5}

1; \pm 5; 10; \pm 10; 5; \pm 1

{(x - 2)}^{5} =

= x^{5} - 10 x^{4} + 40 x^{3} - 80 x^{2} + 80 x - 32

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

= x^{5} + 5 x^{4} + 10 x^{3} + 10 x^{2} + 5 x + 1

now just multiplicate those

who will give x^{5}

1 - 50 + 400 - 800 + 400 - 32 = - 81

Commented by rahul 19 last updated on 27/Mar/18

T h a n k u s i r .

Answered by Tinkutara last updated on 28/Mar/18

{(x^{2} - x - 2)}^{5} = Σ \frac{5!}{a! b! c!} {(x^{2})}^{a} {(- x)}^{b} {(- 2)}^{c}

= Σ \frac{5!}{a! b! c!} {(- 1)}^{b + c} . 2^{c} . x^{2 a + b}

a + b + c = 5, 2 a + b = 5

T h e r e a r e o n l y 3 c a s e s f o r a, b, c :

(1) a = 0, b = 5, c = 0

(2) a = 1, b = 3, c = 1

(3) a = 2, b = 1, c = 2

C o e f f i c i e n t o f x^{2 a + b} i s

= Σ \frac{5!}{a! b! c!} {(- 1)}^{b + c} . 2^{c}

S o w e g e t \frac{5!}{0! 5! 0!} (- 1) + \frac{5!}{3!} \times 2 + \frac{5!}{2! 2!} (- 1) \times 4

= - 1 + 40 - 120 = - 81

Commented by rahul 19 last updated on 28/Mar/18

t h a n k u s o m u c h .