Question

The coefficient of ${{x}^{18}}$ in the product $\left( 1+x \right){{\left( 1-x \right)}^{10}}{{\left( 1+x+{{x}^{2}} \right)}^{9}}$ is?
A. -84
B. 84
C. 126
D. -126

Answer 1

Hint: We will first rewrite the given product expression as, $\left( 1+x \right)\left( 1-x \right){{\left( 1-x \right)}^{9}}{{\left( 1+x+{{x}^{2}} \right)}^{9}}$ and then simplify it and write it as, $\left( 1-{{x}^{2}} \right){{\left( 1-{{x}^{3}} \right)}^{9}}$. We will then use the formula for the ${{r}^{th}}$ term for an expansion ${{\left( a+b \right)}^{n}}$ which is ${}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}$and then use this to proceed.

Complete step-by-step solution:
In the question, we are given a product of expressions which is, $\left( 1+x \right){{\left( 1-x \right)}^{10}}{{\left( 1+x+{{x}^{2}} \right)}^{9}}$ and we have been asked to find the coefficient of ${{x}^{18}}$ in the product. So, we will first consider the given product of expression, $\left( 1+x \right){{\left( 1-x \right)}^{10}}{{\left( 1+x+{{x}^{2}} \right)}^{9}}$ and rearrange it and write it as, $\left( 1+x \right)\left( 1-x \right){{\left( 1-x \right)}^{9}}{{\left( 1+x+{{x}^{2}} \right)}^{9}}$, which can be further written as, $\left( 1-{{x}^{2}} \right){{\left\{ \left( 1-x \right)\left( 1+x+{{x}^{2}} \right) \right\}}^{9}}$. Now, we know that the product of $\left( 1-x \right)$ and $\left( 1+x+{{x}^{2}} \right)$ is $\left( 1-{{x}^{3}} \right)$, so we can rewrite the given expression as follows,
$\left( 1-{{x}^{2}} \right){{\left( 1-{{x}^{3}} \right)}^{9}}$
Now, we will first find the general term of the binomial expansion, ${{\left( 1-{{x}^{3}} \right)}^{9}}$by using the formula for the ${{r}^{th}}$ term for an expansion ${{\left( a+b \right)}^{n}}$, which is given as,
${{T}_{r+1}}={}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}$
Now, as the expansion is ${{\left( 1-{{x}^{3}} \right)}^{9}}$, thus its ${{r}^{th}}$ term will be,
${{T}_{r+1}}={}^{9}{{C}_{r}}{{\left( 1 \right)}^{9-r}}{{\left( -{{x}^{3}} \right)}^{r}}$
Which, on simplification can be written as,
${{T}_{r+1}}={}^{9}{{C}_{r}}{{\left( -1 \right)}^{r}}{{x}^{3r}}$
So, we can now write the expression as,
$\begin{align}
  & \left( 1-{{x}^{2}} \right)\left\{ {}^{9}{{C}_{r}}{{\left( -1 \right)}^{r}}{{x}^{3r}} \right\} \\
 & \Rightarrow {{\left( -1 \right)}^{r}}{}^{9}{{C}_{r}}{{x}^{3r}}-{{\left( -1 \right)}^{r}}{}^{9}{{C}_{r}}{{x}^{3r+2}} \\
\end{align}$
In both terms, the power of x is 3 r and 3 r +2. We have to find the coefficient of ${{x}^{18}}$, so there are two possible causes, the first is when $3r=18$ and the other is $3r+2=18$. The second case is not possible as “r” should be an integer. So, since $3r=18\Rightarrow r=6$.
The coefficient of ${{x}^{3r}}$ was ${{\left( -1 \right)}^{r}}{}^{9}{{C}_{r}}$ and we have obtained the value of $r=6$, so the coefficient will be, ${{\left( -1 \right)}^{6}}{}^{9}{{C}_{6}}\Rightarrow {}^{9}{{C}_{6}}$. To find the value of ${}^{9}{{C}_{6}}$, we have to use the formula of $\begin{align}
  & {}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!} \\
 & \Rightarrow {}^{9}{{C}_{6}}=\dfrac{9!}{\left( 9-6 \right)!6!} \\
 & \Rightarrow {}^{9}{{C}_{6}}=\dfrac{9!}{3!6!} \\
 & \Rightarrow {}^{9}{{C}_{6}}=\dfrac{9\times 8\times 7}{3\times 2\times 1} \\
 & \Rightarrow {}^{9}{{C}_{6}}=3\times 4\times 7 \\
 & \Rightarrow {}^{9}{{C}_{6}}=84 \\
\end{align}$
Therefore, the correct answer is option B.

Note: We can also solve the same question by another method. We can find the products of the given expression and separate out the coefficient of ${{x}^{18}}$ to get the answer, but this method would be very long and tedious, so it is not preferred to be used.