If , then is equal to________. [2026]

Question

If ${(1 - x^{3})}^{10} = \sum_{r = 0}^{10} a_{r} x^{r} {(1 - x)}^{30 - 2 r}$ , then $\frac{9 a_{9}}{a_{10}}$ is equal to________. [2026]

Smart Achievers · Accepted Answer

(30)

Given, ${(1 - x^{3})}^{10} = \sum_{r = 0}^{10} a_{r} x^{r} {(1 - x)}^{30 - 2 r}$

$\Rightarrow {(1 - x^{3})}^{10} = \sum_{r = 0}^{10} a_{r} \frac{x^{r}}{{(1 - x)}^{2 r}} {(1 - x)}^{30}$

$\Rightarrow \frac{((1 - x) {(1 + x + x^{2})}^{10}}{{(1 - x)}^{30}} = \sum_{r = 0}^{10} a_{r} {(\frac{x}{{(1 - x)}^{2}})}^{r}$

$\Rightarrow {(\frac{1 + x + x^{2}}{{(1 - x)}^{2}})}^{10} = \sum_{r = 0}^{10} a_{r} {(\frac{x}{{(1 - x)}^{2}})}^{r}$

$\Rightarrow {(\frac{{(1 - x)}^{2} + 3 x}{{(1 - x)}^{2}})}^{10} = \sum_{r = 0}^{10} a_{r} {(\frac{x}{{(1 - x)}^{2}})}^{r}$

$\Rightarrow {(1 + \frac{3 x}{{(1 - x)}^{2}})}^{10} = \sum_{r = 0}^{10} a_{r} {(\frac{x}{{(1 - x)}^{2}})}^{r}$

Let $\Rightarrow \frac{x}{{(1 - x)}^{2}} = t \Rightarrow {(1 + 3 t)}^{10} = \sum_{r = 0}^{10} a_{r} t^{r}$

For $a_{9} \to Coeff of a_{9} = {}^{10}C_{9} {(3)}^{9}$

For $a_{10} \to Coeff of a_{10} = {}^{10}C_{10} {(3)}^{10}$

$∴ \frac{9 a_{9}}{a_{10}} = \frac{9 \times {}^{10}C_{9} 3^{9}}{{}^{10}C_{10} 3^{10}} = 3 \times 10 = 30$

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