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} $ .