# If the coefficients of x3 and x4 in the expansion of (1 + ax + bx2) (1 - 2x)18 in powers of x are both zero, then (a, b) is equal to : Option 1) Option 2) Option 3) Option 4)

G gaurav

As we have learned

Expression of Binomial Theorem -

$\left ( x+a \right )^{n}= ^{n}\! c_{0}x^{n}a^{0}+^{n}c_{1}x^{n-1}a^{1}+$$^{n}c_{2}x^{n-2}a^{2}x-----^{n}c_{n}x^{0}a^{n}$

- wherein

for n  +ve integral .

$(1+ax +bx^2)(1-2x)^{18} = (1+ax +bx^2) (^{18 }C_0-^{18}C_1(2x)+^{18}C_2(2x^2)....)$

coeff of  $x^3 = - ^{18}C_3(2)^3+ ^{18}C_2(2)^2a - ^{18}C_1(2)b= 0$

coeff of  $x^4 = - ^{18}C_4(2)^4+ ^{18}C_3(2)^3a + ^{18}C_2(2)^2b= 0$

$\Rightarrow \frac{-18 \times 17 \times 16}{3 \times 2} \times 8 + \frac{18 \times 17}{2}\times 4a -18 \times 2b = 0$ and

$\Rightarrow \frac{18 \times 17 \times 16\times 15}{4\times 3 \times 2} \times 16 + \frac{18 \times 17\times 16}{6}\times 8a -\frac{18\times 17}{2} \times 4b = 0$

$\Rightarrow 12 (-544 + 51 a -3b )= 0$

and  $12 \times 17 (240 - 32 a +3b ) = 0$

$51 a -3b -544 = 0\: \: \: \& \: \: \: -32 a +3b +240 = 0$

a = 16

b = 272/3

Option 1)

Option 2)

Option 3)

Option 4)

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