If (1-x^3)^10 = _r=0^10 a_r,x^r(1-x)^30-2r, then 9a_9a_10 is equal to ____.

Question

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

Answer: 30

Accepted Answer

Write $1-x^3=(1-x)(1+x+x^2)$, so $(1-x^3)^{10}=(1-x)^{10}(1+x+x^2)^{10}$. Comparing with the given form,

$$(1+x+x^2)^{10} = \sum_{r=0}^{10} a_r x^r(1-x)^{20-2r}$$

Now write $1+x+x^2 = (1-x)^2+3x$, so

$$(1+x+x^2)^{10} = \big((1-x)^2+3x\big)^{10} = \sum_{r=0}^{10}\binom{10}{r}(3x)^r\big((1-x)^2\big)^{10-r} = \sum_{r=0}^{10}\binom{10}{r}3^r x^r(1-x)^{20-2r}$$

Hence $a_r=\binom{10}{r}3^r$. So

$$a_9=\binom{10}{9}3^9=10\cdot 3^9,\qquad a_{10}=\binom{10}{10}3^{10}=3^{10}$$

$$\frac{9a_9}{a_{10}} = 9\cdot\frac{10\cdot 3^9}{3^{10}} = 9\cdot\frac{10}{3}=30$$

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