Question

The coefficient of $x^{10}$ in $(1-x^2)(1-x^3)^9$ is:

• A. ${}^9C_4$
• B. $-{}^9C_6$
• C. $-{}^9C_4$
• D. ${}^9C_6$
• E. 0

Hint:

In $(1-x^k)^n$, all powers of $x$ must be multiples of $k$.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
We need to find the coefficient of a specific power of \( x \) in the expansion of a binomial expression. We will use the binomial theorem.
Step 2: Key Formula or Approach:
The general term, \( T_{r+1} \), in the binomial expansion of \( (a+b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \] For the expression \( (1 - x^3)^{15} \), we have \( a=1, b = -x^3 \), and \( n=15 \).
Step 3: Detailed Explanation:
Let's find the general term for the expansion of \( (1 - x^3)^{15} \).
\[ T_{r+1} = \binom{15}{r} (1)^{15-r} (-x^3)^r \] Simplifying this, we get:
\[ T_{r+1} = \binom{15}{r} (-1)^r (x^3)^r \] \[ T_{r+1} = \binom{15}{r} (-1)^r x^{3r} \] We are looking for the coefficient of \( x^{10} \). To find this, we must set the exponent of \( x \) in the general term equal to 10.
\[ 3r = 10 \] Now, we solve for \( r \):
\[ r = \frac{10}{3} \] For the binomial expansion, the index \( r \) must be a non-negative integer (\( r \in \{0, 1, 2, \dots, 15\} \)).
Since \( r = \frac{10}{3} \) is not an integer, it means that there is no term in the expansion that contains \( x^{10} \).
The powers of \( x \) that appear in the expansion are all multiples of 3 (e.g., \( x^0, x^3, x^6, x^9, x^{12}, \dots \)). Since 10 is not a multiple of 3, the term \( x^{10} \) does not exist in this expansion.
Step 4: Final Answer:
The coefficient of a term that does not exist is 0.