If the coefficients of x3 and x4 in the expansion of (1+ax+bx2) (1

Question

If the coefficients of

x^{3}

and

x^{4}

in the expansion of

(1 + a x + b x^{2}

)

(1 - 2 x)^{18}

in powers of

x

are both zero, then

(a, b)

is equal to

Options:

A . (16,2513)

B . (14,2513)

C . (14,2723)

D . (16,2723)

Answer: Option D
:
D

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

Coefficient of

x^{3} : (- 2)^{3}^{18} C_{3} + a (- 2)^{2}^{18} C_{2} + b (- 2)^{18} C_{1} = 0

\frac{4 \times (17 \times 16)}{(3 \times 2)} - 2 a \times \frac{17}{2} + b = 0

----- (1)
Coefficient of

x^{4} : (- 2)^{4}^{18} C_{4} + a (- 2)^{3}^{18} C_{3} + b (- 2)^{2}^{18} C_{2} = 0

(4 \times 20) - 2 a \times \frac{16}{3} + b = 0

---- (2)
From equations (1) and (2), we get

4 (\frac{17 \times 8}{3} - 20) + 2 a (\frac{16}{3} - \frac{17}{2}) = 0

\Rightarrow

a = 16

\Rightarrow b = \frac{2 \times 16 \times 16}{3} - 80 = \frac{272}{3}

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