If-1-ax-bx-2-1-2x-18-can-be-expanded-using-binomial-theorem-in-ascending-power-of-x-Determine-the-value-of-a-and-b-if-the-coefficient-of-x-3-and-x-4-are-both-zero-

Question Number 40523 by scientist last updated on 23/Jul/18

I f (1 + a x + {b x}^{2}) {(1 - 2 x)}^{18} c a n b e e x p a n d e d u s i n g

b i n o m i a l t h e o r e m i n a s c e n d i n g p o w e r o f x . D e t e r m i n e

t h e v a l u e o f a a n d b, i f t h e c o e f f i c i e n t o f x^{3} a n d x^{4} a r e b o t h z e r o .

Commented by tanmay.chaudhury50@gmail.com last updated on 23/Jul/18

(1 + a x + {b x}^{2}) (1 - 18 C_{1} . 2 x + 18 C_{2} . 4 x^{2} - 18 C_{3} . 8 x^{3}

+ 18 C_{4} . 16 x^{4} + \dots)

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

x^{4} (18 C_{4} . 16 - a . 18 C_{3} . 8 + b . 18 C_{2} . 4)

t h e t e r m c o n t a i n i n g x^{3}

x^{3} (- 18 C_{3} . 8 + a . 18 C_{2} . 4 - b . 18 C_{1} . 2)

i f c o e f f i c i e n t o f x^{3} a n d x^{4} a r e z e r o t h e n

a . 18 C_{2} . 4 - b . 18 C_{1} . 2 = 18 C_{3} . 16 i s t e q n

a . 18 C_{3} . 8 - b . 18 C_{2} . 4 = 18 C_{4} . 16 2 n d e q n

t o s o l v e \dots

18 C_{1} = 18 18 C_{2} = \frac{18 \times 17}{2} = 153

18 C_{3} = \frac{18 \times 17 \times 16}{3 \times 2} = \frac{153 \times 16}{3} = 816

18 C_{4} = \frac{18 \times 17 \times 16 \times 15}{4 \times 3 \times 2} = \frac{816 \times 15}{4} = 3060

i s t e q n

612 a - 36 b = 816 \times 16

816 \times 8 a - 153 \times 4 b = 3060 \times 16

i t s a b i g b i g n u m b e r s o s o l v e p l s \dots

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Jul/18

(1 - 18 C_{1} . 2 x + 18 C_{2} . {(2 x)}^{2} - 18 C_{3} . {(2 x)}^{3} + \dots .)

t e r m s c o n t a i n i n g x^{3}

= - 18 C_{3} {(2 x)}^{3} + a x . 18 C_{2} . {(2 x)}^{2} + {b x}^{2} . (- 18 C_{1} (2 x)

= x^{3} {- 18 C_{3} . 8 + a . 4.18 C_{2} + b . - 18 C_{3} . 2}

p l s p o s t c o m p l e t e q u e s t i o n \dots

Commented by MJS last updated on 23/Jul/18

also the constant factor of x^{4} is zero, so just

continue!

Commented by tanmay.chaudhury50@gmail.com last updated on 23/Jul/18

o k s i r \dots