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JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Binomial Theorem

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

Answer: 30

Solution

Write

1 - x^{3} = (1 - x)^{3} + 3 x (1 - x)

. Then

(1 - x^{3})^{10} = [(1 - x)^{3} + 3 x (1 - x)]^{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 with the given form,

a_{r} = (r 10) 3^{r}

, so

a_{9} = (9 10) 3^{9} = 10 \cdot 3^{9}

and

a_{10} = 3^{10}

. Hence

\frac{9 a _{9}}{a _{10}} = \frac{9 \cdot 10 \cdot 3 ^{9}}{3 ^{10}} = \frac{90}{3} = 30

.

JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Integer answer

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