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

Concept: Use the identity \[ 1-x^3=(1-x)(1+x+x^2) \] and then apply the binomial theorem. Step 1: {Rewrite the expression.} \[ (1-x^3)^{10} \] \[ =(1-x)^{10}(1+x+x^2)^{10} \] Thus \[ (1+x+x^2)^{10}=\sum_{r=0}^{10}a_r x^r (1-x)^{20-2r} \] Step 2: {Compare highest power coefficients.} For \(x^{10}\): \[ a_{10}=\binom{10}{10}=1 \] For \(x^9\): \[ a_9=\binom{10}{9}=10 \] Step 3: {Compute the required ratio.} \[ \frac{9a_9}{a_{10}} \] \[ =\frac{9\times10}{1} \] \[ =90 \]