JEE Main 2025 Jan 28 Shift 2, Mathematics Q18: Properties Of Binomial Coefficient

JEE Main2025Jan 28, Shift 2Mathematics

Let the coefficients of three consecutive terms $T_{r}$ , $T_{r + 1}$ and $T_{r + 2}$ in the binomial expansion of $(a + b)^{12}$ be in a G.P. and let $p$ be the number of all possible values of $r$ . Let $q$ be the sum of all rational terms in the binomial expansion of $(43 + 34)^{12}$ . Then $p + q$ is equal to :

A
$283$
B
$295$
C
$287$
D
$299$

Solution

GP condition on consecutive binomial coefficients: $\frac{^{12} C _{r + 1}}{^{12} C _{r}} = \frac{^{12} C _{r + 2}}{^{12} C _{r + 1}}$ .

Using $\frac{^{n} C _{k + 1}}{^{n} C _{k}} = \frac{n - k}{k + 1}$ , the equation becomes $(13 - r) (r + 1) = (12 - r) (r) + \dots$ which simplifies to $13 = 0$ , impossible. So $p = 0$ .

General term of $(3^{1/4} + 4^{1/3})^{12}$ : $T_{k + 1} =^{12} C_{k} 3^{(12 - k) /4} 4^{k /3}$ . It is rational iff $4 ∣ (12 - k)$ and $3 ∣ k$ , i.e. $k \in {0, 12}$ .

$k = 0$ : $^{12} C_{0} \cdot 3^{3} = 27$ . $k = 12$ : $^{12} C_{12} \cdot 4^{4} = 256$ . So $q = 27 + 256 = 283$ .

$∴ p + q = 0 + 283 = 283$ .

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JEE Main 2025 Jan 28 Shift 2, Mathematics Q18: Properties Of Binomial Coefficient

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