Binomial Theorem Ques 24
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) $(16, \frac{251}{3})$
(b) $(14, \frac{251}{3})$
(c) $(14, \frac{272}{3})$
(d) $(16, \frac{272}{3})$
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Answer:
Correct Answer: 24.(d)
Solution:
Formula:
Properties of binomial theorem:
- To find the coefficient of $x^{3}$ and $x^{4}$, use the formula of coefficient of $x^{r}$ in $(1-x)^{n}$ is $(-1)^{r} \mathrm{C}_{r}$ and then simplify.
In expansion of $\left(1+a x+b x^{2}\right)(1-2 x)^{18}$.
Coefficient of $x^{3}=$ Coefficient of $x^{3}$ in $(1-2 x)^{18}$
$ \begin{gathered} \text { + Coefficient of } x^{2} \text { in } a(1-2 x)^{18} \\ \text { + Coefficient of } x \text { in } b(1-2 x)^{18} \\ ={ }^{18} C_{3} \cdot 2^{3}+a{ }^{18} C_{2} \cdot 2^{2}-b{ }^{18} C_{1} \cdot 2 \end{gathered} $
Given, coefficient of $x^{3}=0$
$\Rightarrow { }^{18} C_{3} \cdot 2^{3}+a^{18} C_{2} \cdot 2^{2}-b{ }^{18} C_{1} \cdot 2=0 $
$\Rightarrow -\frac{18 \times 17 \times 16}{3 \times 2} \cdot 8+a \cdot \frac{18 \times 17}{2} \cdot 2^{2}-b \cdot 18 \cdot 2=0 $
$\Rightarrow 17 a-b=\frac{34 \times 16}{3}$ $\quad$ …….(i)
Similarly, coefficient of $x^{4}=0$
$ \begin{array}{rlrl} \Rightarrow & { }^{18} C_{4} \cdot 2^{4}-a \cdot{ }^{18} C_{3} 2^{3}+b \cdot{ }^{18} C_{2} \cdot 2^{2}=0 \\ \therefore & 32 a-3 b =240 \quad …….(ii) \end{array} $
On solving Eqs. (i) and (ii), we get
$ a=16, \quad b=\frac{272}{3} $
BETA
