Question

If \[ (1-x^3)^{10}=\sum_{r=0}^{10}a_r x^r(1-x)^{30-2r}, \] then \( \dfrac{9a_9}{a_{10}} \) is equal to ________.

Answer 1

Correct Answer: 90

Step 1: Understanding the Question
We are given an identity that must hold for all \(x\). The left side is a binomial expression, and the right side is a sum involving coefficients \(a_r\). We need to find these coefficients (or at least their ratio) to compute the final expression. The structure of the sum on the right side suggests a substitution might simplify the identity.
Step 2: Key Formula or Approach
1. Manipulate the given identity into a recognizable form, likely a standard binomial expansion.
2. Identify a suitable substitution to simplify the expression.
3. Use the binomial theorem: \( (1+z)^n = \sum_{r=0}^{n} \binom{n}{r} z^r \).
4. Compare coefficients to find an expression for \(a_r\).
Step 3: Detailed Explanation
The given identity is: \[ (1-x^3)^{10} = \sum_{r=0}^{10} a_r x^r (1-x)^{30-2r} \] Let's try to make the terms in the sum look like powers of a single variable. We can divide by \( (1-x)^{30} \): \[ \frac{(1-x^3)^{10}}{(1-x)^{30}} = \sum_{r=0}^{10} a_r \frac{x^r (1-x)^{30-2r}}{(1-x)^{30}} \] \[ \frac{((1-x)(1+x+x^2))^{10}}{(1-x)^{30}} = \sum_{r=0}^{10} a_r \frac{x^r}{(1-x)^{2r}} \] \[ \frac{(1-x)^{10}(1+x+x^2)^{10}}{(1-x)^{30}} = \sum_{r=0}^{10} a_r \left( \frac{x}{(1-x)^2} \right)^r \] \[ \left( \frac{1+x+x^2}{(1-x)^2} \right)^{10} = \sum_{r=0}^{10} a_r \left( \frac{x}{(1-x)^2} \right)^r \] Let's define a new variable \( y = \frac{x}{(1-x)^2} \). The identity becomes: \[ \left( \frac{1+x+x^2}{(1-x)^2} \right)^{10} = \sum_{r=0}^{10} a_r y^r \] Now, we need to express the term on the left in terms of \(y\). \[ \frac{1+x+x^2}{(1-x)^2} = \frac{1-2x+x^2+3x}{1-2x+x^2} = \frac{(1-x)^2+3x}{(1-x)^2} = 1 + \frac{3x}{(1-x)^2} = 1+3y \] Substituting this back into the equation: \[ (1+3y)^{10} = \sum_{r=0}^{10} a_r y^r \] This is the binomial expansion of \( (1+3y)^{10} \). According to the binomial theorem: \[ (1+3y)^{10} = \sum_{r=0}^{10} \binom{10}{r} (3y)^r = \sum_{r=0}^{10} \binom{10}{r} 3^r y^r \] By comparing the coefficients of \(y^r\) in both summations, we find: \[ a_r = \binom{10}{r} 3^r \] Now we can compute \( a_9 \) and \( a_{10} \). \[ a_9 = \binom{10}{9} 3^9 = 10 \cdot 3^9 \] \[ a_{10} = \binom{10}{10} 3^{10} = 1 \cdot 3^{10} \] Finally, we calculate the required expression: \[ \frac{9a_9}{a_{10}} = \frac{9 \cdot (10 \cdot 3^9)}{1 \cdot 3^{10}} = \frac{9 \cdot 10}{3} = 3 \cdot 10 = 30 \] Step 4: Final Answer
The value of \( \frac{9a_9}{a_{10}} \) is 30.