Question

Find the coefficient of ${{x}^{18}}$ in ${{\left( a{{x}^{4}}-bx \right)}^{9}}$.

Answer 1

Hint: We will start by using the binomial theorem to expand ${{\left( a{{x}^{4}}-bx \right)}^{9}}$. Then we will find the coefficient of the terms in which has ${{x}^{18}}$ as a variable term for this. We will use the fact that if ${{\left( a+b \right)}^{n}}$ is an binomial expression then its binomial term can be represented as ${}^{n}{{C}_{r}}{{a}^{r}}{{b}^{n-r}}$.

Complete step by step solution:
Now, we have to find the coefficient of ${{x}^{18}}$ in ${{\left( a{{x}^{4}}-bx \right)}^{9}}$.
Now, we know that the binomial expansion of the expression ${{\left( a+b \right)}^{n}}$ is,
${{\left( a+b \right)}^{n}}={}^{n}{{C}_{0}}{{a}^{0}}{{b}^{n}}+{}^{n}{{C}_{1}}{{a}^{1}}{{b}^{n-1}}+{}^{n}{{C}_{2}}{{a}^{2}}{{b}^{n-2}}+...........+{}^{n}{{C}_{n}}{{a}^{n}}{{b}^{0}}$
We can see that the ${{r}^{th}}$ term of such series is ${}^{n}{{C}_{r}}{{a}^{r}}{{b}^{n-r}}$.
Now, we have the expression as ${{\left( a{{x}^{4}}-bx \right)}^{9}}$.
Now, we have to find the term which has ${{x}^{18}}$ as a variable term.
Now, we can write the ${{r}^{th}}$ term of the expression as ${}^{n}{{C}_{r}}{{\left( a{{x}^{4}} \right)}^{r}}{{\left( -bx \right)}^{n-r}}$.
$\begin{align}
  & tn={}^{tn}{{C}_{r}}{{a}^{r}}{{x}^{4r}}\times {{\left( -b \right)}^{tn-r}}{{x}^{tn-r}} \\
 & ={}^{n}{{C}_{r}}{{a}^{r}}{{\left( -b \right)}^{n-r}}\times {{x}^{4r+n-r}} \\
 & ={}^{n}{{C}_{r}}{{a}^{r}}{{\left( -b \right)}^{n-r}}\times {{x}^{n+3r}} \\
\end{align}$
Now, we have to find the coefficient of ${{x}^{18}}$. Therefore, on comparing variable with the tn term we have,
$\begin{align}
  & \Rightarrow {{x}^{n+3r}}={{x}^{18}} \\
 & \Rightarrow n+3r=18 \\
\end{align}$
Now, n = 9 for ${{\left( a{{x}^{4}}-bx \right)}^{9}}$. So, we have,
$\begin{align}
  & \Rightarrow 3r+9=18 \\
 & \Rightarrow 3r=9 \\
 & \Rightarrow r=3 \\
\end{align}$
So, we have the 3rd term as,
$\begin{align}
  & {{t}_{3}}={}^{9}{{C}_{3}}{{a}^{3}}{{\left( -b \right)}^{6}}{{x}^{18}} \\
 & ={}^{9}{{C}_{3}}{{a}^{3}}{{b}^{6}}{{x}^{18}} \\
\end{align}$
So, the coefficient of ${{x}^{18}}$ is ${}^{9}{{C}_{3}}{{a}^{3}}{{b}^{6}}$. Further we know that ${}^{n}{{C}_{3}}=\dfrac{n\left( n-1 \right)\left( n-2 \right)}{6}$. So, we have the coefficient of ${{x}^{18}}$ as $\dfrac{9\times 8\times 7}{6}{{a}^{3}}{{b}^{3}}=84{{a}^{3}}{{b}^{3}}$.

Note: It is important to note that we have used a fact that the ${{r}^{th}}$ term in the binomial expansion of ${{\left( a+b \right)}^{n}}$ is ${}^{n}{{C}_{r}}{{a}^{r}}{{b}^{n-r}}$. Also, we have used a fact that ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ or for r = 3 it can be remembered as ${}^{n}{{C}_{3}}=\dfrac{n\left( n-1 \right)\left( n-2 \right)}{6}$.