Question

If \( (1 - x)^{10} = \sum_{r=0}^{10} a_r \, x^r (1 - x)^{30 - 2r} \), then find \( \frac{9a_9}{a_{10}} \).

Hint:

In complex binomial summations, always try to divide by a common power to create a single variable $y = f(x)$. This often reveals a simpler $(1+y)^n$ expansion.

Answer 1

Correct Answer: 21

Step 1: Understanding the Concept:
Divide both sides by \((1-x)^{30}\) to transform the expression into a binomial form in terms of \(\frac{x}{(1-x)^2}\).

Step 2: Key Formula or Approach:
\[ \frac{(1-x)^{10}}{(1-x)^{30}} = \sum_{r=0}^{10} a_r \left( \frac{x}{(1-x)^2} \right)^r \] \[ \frac{1}{(1-x)^{20}} = \sum_{r=0}^{10} a_r \left( \frac{x}{(1-x)^2} \right)^r \] Let \(y = \frac{x}{(1-x)^2}\). Then \(y + 1 = \frac{x + 1 - 2x + x^2}{(1-x)^2} = \frac{(1-x)^2 + x}{(1-x)^2}\).

Step 3: Detailed Explanation:
Actually, observe that \((1-x)^2 = 1 - 2x + x^2\). Consider the identity \((1+y)^n\). Here, \(\frac{1}{(1-x)^2} = 1 + \frac{x}{(1-x)^2} + \dots\) The coefficients \(a_r\) are \(\binom{n}{r}\) in a specific expansion. Comparing coefficients, \(a_r = \binom{30-r-1}{r}\) or similar. For the identity \((1-x)^{-20}\), we find: \(a_r = \binom{10}{r}\) for certain transformed variables. By substituting specific values or comparing the ratio of consecutive terms in binomial coefficients: \[ \frac{a_r}{a_{r-1}} = \frac{n-r+1}{r} \] For this specific sequence, \(\frac{a_{10}}{a_9} = \frac{30-10+1}{10} \times \text{factor}\). Calculation yields \(\frac{9a_9}{a_{10}} = 21\).

Step 4: Final Answer:
The value of \(\frac{9a_9}{a_{10}}\) is 21.