JEE Main 2026 Apr 6 Shift 2, Mathematics Q22: Introduction to Binomial Expansion

JEE Main2026Apr 6, Shift 2Mathematics

If $(1 - x^{3})^{10} = r = 0 \sum 10 a_{r} x^{r} (1 - x)^{30 - 2 r}$ , then $\frac{9 a _{9}}{a _{10}}$ is equal to ______.

Your answer

Solution

Use the identity $(1 - x)^{3} = 1 - 3 x + 3 x^{2} - x^{3}$ , so $1 - x^{3} = (1 - x)^{3} + 3 x (1 - x)$ .

Raise to the 10th power and apply the binomial theorem:

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

Comparing, $a_{r} = (r 10) 3^{r}$ .

$\frac{a _{9}}{a _{10}} = \frac{( 9 10 ) 3 ^{9}}{( 10 10 ) 3 ^{10}} = \frac{10}{3}$ . So $\frac{9 a _{9}}{a _{10}} = 9 \cdot \frac{10}{3} = 30$ .

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