Question

Find the coefficient of ${x^{32}}$ in the expansion of ${\left( {{x^4} - \dfrac{1}{{{x^3}}}} \right)^{15}}$
A. ${}^{ - 15}{C_3}$
B.${}^{15}{C_4}$
C. ${}^{ - 15}{C_5}$
D. ${}^{15}{C_2}$

Answer 1

Hint : In this problem regarding expansion of the binomial, the formula for the general term or r-th term is used. The general or r-th term in the expansion of ${\left( {a + b} \right)^n}$ is ${T_{r + 1}} = {}^n{C_r} \times {\left( a \right)^{n - r}} \times {b^r}$ . The value of a and b should be compared with the binomial expression given in the question and should be substituted in the expression of the general term. After that power x should be equated with $32$ to determine the value of $r$ .

Complete step-by-step answer :
The given binomial expression is
$E = {\left( {{x^4} - \dfrac{1}{{{x^3}}}} \right)^{15}} \cdots \left( 1 \right)$
We are required to determine the coefficient of ${x^{32}}$ in the given expansion.
The general term in the expansion of ${\left( {a + b} \right)^n}$ is given by,
${T_{r + 1}} = {}^n{C_r} \times {\left( a \right)^{n - r}} \times {b^r} \cdots \left( 2 \right)$
On comparing equation (1) with ${\left( {a + b} \right)^n}$ , we can say that
$a = {x^4}$ , $b = \dfrac{1}{{{x^3}}}$ and $n = 15.$
Substituting the value of a ,b and n in equation (2), we get
${T_{r + 1}} = {}^{15}{C_r} \times {\left( {{x^4}} \right)^{15 - r}} \times {\left( {\dfrac{1}{{{x^3}}}} \right)^r} \cdots \left( 3 \right)$
Simplifying equation (3), we get
$\Rightarrow {T_{r + 1}} = {}^{15}{C_r} \times {\left( {{x^4}} \right)^{4 - r}} \times {\left( {{x^{ - 3}}} \right)^r} \\\ {T_{r + 1}} = {}^{15}{C_r} \times {\left( x \right)^{60 - 4r}} \times {\left( x \right)^{ - 3r}} \cdots \left( 4 \right) \;$
When the base is the same then powers are added. Here x is the base and $\left( {60 - 7r} \right)$ and $- 3r$ are the powers, which are to be added.
Solving equation (4), we get
$\Rightarrow {T_{r + 1}} = {}^{15}{C_r} \times {x^{60 - 4r - 3r}} \\\ \Rightarrow {T_{r + 1}} = {}^n{C_r} \times {x^{60 - 7r}} \cdots \left( 5 \right) \;$
Since we are required to determine the coefficient of ${x^{32}}$ . Therefore the value of $60 - 7r$ is equal to $32$ i.e.,
$60 - 7r = 32 \cdots \left( 6 \right)$
Solving equation (6) for r, we get
$\Rightarrow 60 - 7r = 32 \\\ \Rightarrow 7r = 60 - 32 \\\ \Rightarrow 7r = 28 \\\ \Rightarrow r = 4 \;$
Substitute the value of $r = 4$ in equation (6), we get

$$\Rightarrow {T_{4 + 1}} = {}^{15}{C_4} \times {\left( x \right)^{60 - 7 \times 4}} \\\ \Rightarrow {T_5} = {}^{15}{C_4} \times {\left( x \right)^{32}} \cdots \left( 7 \right) \;$$

It is clear from equation (7) that the coefficient of ${x^{32}}$ is ${}^{15}{C_4}$ in the expansion of ${\left( {{x^4} - \dfrac{1}{{{x^3}}}} \right)^{15}}$ .
Thus, the correct option is (B).
So, the correct answer is “Option B”.

Note : The formula for general term in the expansion of ${\left( {a + b} \right)^n}$ is ${T_{r + 1}} = {}^n{C_r} \times {\left( a \right)^{n - r}} \times {b^r}$ and it should be clear in mind. The important thing is to get the values of a and b present in the general term ${T_{r + 1}} = {}^n{C_r} \times {\left( a \right)^{n - r}} \times {b^r}$ .