Binomial Theorem 1 Question 1
1.
The coefficient of $x^{18}$ in the product $(1+x)(1-x)^{10}\left(1+x+x^{2}\right)^{9}$ is
(2019 Main, 12 April I)
(a) 84
(b) -126
(c) -84
(d) 126
Show Answer
Answer:
Correct Answer: 1. (a)
Solution:
- Given expression is
$ \begin{aligned} (1+x) & (1-x)^{10}\left(1+x+x^{2}\right)^{9} \\ \quad= & (1+x)(1-x)\left[(1-x)\left(1+x+x^{2}\right)\right]^{9} \\ \quad= & \left(1-x^{2}\right)\left(1-x^{3}\right)^{9} \end{aligned} $
Now, coefficient of $x^{18}$ in the product
$ \begin{aligned} & \qquad(1+x)(1-x)^{10}\left(1+x+x^{2}\right)^{9} \\ & =\text { coefficient of } x^{18} \text { in the product }\left(1-x^{2}\right)\left(1-x^{3}\right)^{9} \\ & =\text { coefficient of } x^{18} \text { in }\left(1-x^{3}\right)^{9} - \text {coefficient of } x^{16} \text { in }\left(1-x^{3}\right)^{9} \end{aligned} $
Since, $(r+1)^{\text {th }}$ term in the expansion of
$ \left(1-x^{3}\right)^{9} \text { is }{ }^{9} C_{r}\left(-x^{3}\right)^{r}={ }^{9} C_{r}(-1)^{r} x^{3 r} $
Now, for $x^{18}, 3 r=18 \Rightarrow r=6$
and for $x^{16}, 3 r=16$
$\Rightarrow \quad r=\frac{16}{3} \notin N$.
$\therefore$ Required coefficient is ${ }^{9} C_{6}=\frac{9 !}{6 ! 3 !}=\frac{9 \times 8 \times 7}{3 \times 2}=84$
