Q21 / 50 · Manipal (MET) 2023 · Mathematics
If the coefficients of $x^3$ and $x^4$ in the expansion of $\left(1+a x+b x^2\right)(1-2 x)^{18}$ in powers of $x$ are both zero, then $(a, b)$ is equal to
- A$\left(14, \frac{272}{3}\right)$
- Correct answerB$\left(16, \frac{272}{3}\right)$
- C$\left(16, \frac{251}{3}\right)$
- D$\left(14, \frac{251}{3}\right)$
Solution
In the expansion of $\left(1+a x+b x^2\right)(1-2 x)^{18}$ coefficient of $x^3$ in $\left(1+a x+b x^2\right)(1-2 x)^{18}$
$=$ Coefficient of $x^3$ in $(1-2 x)^{18}+$ coefficient of $x^2$ in $a(1-2 x)^{18}+$ coefficient of $x$ in $b(1-2 x)^{18}$.
$=-{ }^{18} C_3 \cdot 2^3+a^{18} C_2 \cdot 2^2-b^{18} C_4 \cdot 2$
$\because \quad-{ }^{18} C_3 \cdot 2^3+a^{18} C_2 \cdot 2^2-b^{18} C_4 \cdot 2=0$
$\Rightarrow \frac{18 \times 17 \times 16}{3 \times 2} \cdot 8+a \cdot \frac{18 \times 17}{2} 2^2-b \cdot 18 \cdot 2=0$
$\Rightarrow 17 a-b=\frac{34 \times 16}{3}\ldots$ (i)
Similarly, coefficient of $x^4$
${ }^{18} C_4 \cdot 2^4-a \cdot{ }^{18} C_3 2^3+b \cdot{ }^{18} C_2 \cdot 2^2=0$
$\therefore \quad 32 a-32 b=240\ldots$ (ii)
From Eqs. (i) and (ii), we get
$a=16$ and $b=\frac{272}{3}$
Hence, the correct answer is option: 'B'.