JEE Main · 2014 · 06 Apr 2014 · Offline · Binomial Theorem

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

Answer: (D)

Solution

Expand

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

and collect the required powers, remembering that the coefficient of

x^{k}

(1 - 2 x)^{18}

(k 18) (- 2)^{k}

. The coefficient of

x^{3}

(3 18) (- 2)^{3} + a (2 18) (- 2)^{2} + b (1 18) (- 2) = - 6528 + 612 a - 36 b

, and setting it to zero gives

51 a - 3 b = 544

. The coefficient of

x^{4}

(4 18) (- 2)^{4} + a (3 18) (- 2)^{3} + b (2 18) (- 2)^{2} = 48960 - 6528 a + 612 b

, and setting it to zero gives

544 a - 51 b = 4080

. Solving the pair,

323 a = 5168

a = 16

, and then

816 - 3 b = 544

b = \frac{272}{3}

. Thus

(a, b) = (16, \frac{272}{3})

JEE Main · 2014 · 06 Apr 2014 · Offline · Single correct option