First slide

Binomial theorem for positive integral Index

Question

If coefficient 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 zeros, then $(a, b)$ is equal to

Moderate

A

$(14, \frac{251}{3})$
B

$(14, \frac{272}{3})$
C

$(16, \frac{272}{3})$
D

$(16, \frac{251}{3})$

Solution

$S = (1 + a x + b x^{2}) (1 - 2 x)^{18}$
$\begin{array}{l} = (1 + a x + b x^{2}) [1 +^{18} C_{1} (- 2 x) + \\ ^{18} C_{2} (- 2 x)^{2} +^{18} C_{3} (- 2 x)^{3} +^{18} C_{4} (- 2 x)^{4} + \dots] \end{array}$
Coefficient of $x^{3}$ in the expansion of S is $^{18} C_{3} (- 2)^{3} +$ $a (^{18} C_{2} (- 2)^{2}) +^{18} C_{1} (- 2) b = 0$
Divide by $^{18} C_{1} (- 2)$ to obtain
$\frac{544}{3} - 17 a + b = 0$ (1)
Similarly, coefficient of $x^{4}$ is
$^{18} C_{4} (- 2)^{4} + a (^{18} C_{3} (- 2)^{3}) +^{18} C_{2} (- 2)^{2} b = 0$
Divide by $^{18} C_{2} (- 2)^{2}$ to obtain
$80 - \frac{32}{3} a + b = 0$ (2)
Subtract (2) from (1) to obtain
$\frac{304}{3} - \frac{19}{3} a = 0 \Rightarrow a = 16$
From $b = 17 \times 16 - \frac{544}{3} = \frac{272}{3}$

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