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Q.
If the coefficients of $ x^3$ and $x^4 $in the expansion of $(1 + ax + bx^2) (1 - 2x)^{18}$ in powers of $x$ are both zero, then $(a, b)$ is equal to
Binomial Theorem
Solution:
To find the coefficient of $x^3$ and $x^4$, use the formula of coefficient of $x^r$ in $(1 - x)^n$ is $( - 1 )^r$ $ {}^{n}C_r$and then simplify
. In expansion of $(1 + ax + bx^2)(1 - 2x)^{18}$
Coefficient of $x^3$ = Coefficient of $x^3$ in $(1 - 2x)^{18}$
$ +$ Coefficient of $x^2$ in a $ (1 -2 x )^{18}$
$ +$ Coefficient of $x$ in $b(1 - 2x)^{18}$
$ \, = - ^{18}C_3.2^3 +a \,{}^{18}C_a.2^2-b^{18}C_1.2$
Given, coefficient of $x^3 = 0$
$\Rightarrow {}^{18}C_3.2^3+a \,{}^{18}C_2.2^2-b \,{}^{18}C_1.2=0$
$\Rightarrow -\frac{18 \times \, 17 \times \, 16}{3 \times\, 2}.8+a.\frac{18 \times 17}{2} 2^2-b.18.2=0 $
$\Rightarrow 17 a-b=\frac{34\times 16}{3} $....(i)
Similarly, coefficient of $x^4 = 0$
$\Rightarrow {}^{18}C_4.2^4-a.\,{}^{18}C_3 2^3+b. \,{}^{18}C_2.2^2=0$
$\therefore 32a-3b=240$.......(ii)
On solving Eqs. (i) and (ii), we get
$ a=16,b=\frac{272}{3}$