Binomial Theorem – Properties of Binomial Coefficients – JEE Main 21 Jan 2026 Shift 2

Solution:

Let the given product be $P$. The general term of the product can be written as:
$$ T_r = \frac{1}{^{15}C_r} + \frac{1}{^{15}C_{r+1}} \quad \text{for } r = 0, 1, \dots, 12 $$

We simplify the general term:
$$ T_r = \frac{^{15}C_{r+1} + ^{15}C_r}{^{15}C_r \cdot ^{15}C_{r+1}} $$

Using the property $^nC_r + ^nC_{r+1} = ^{n+1}C_{r+1}$:
$$ T_r = \frac{^{16}C_{r+1}}{^{15}C_r \cdot ^{15}C_{r+1}} $$

We use the ratio property $\frac{^nC_k}{^{n-1}C_{k-1}} = \frac{n}{k}$. Here, $\frac{^{16}C_{r+1}}{^{15}C_r} = \frac{16}{r+1}$.
$$ T_r = \frac{16}{r+1} \cdot \frac{1}{^{15}C_{r+1}} $$

Now, using the property $^nC_k = \frac{n}{k} \cdot ^{n-1}C_{k-1}$ for the denominator term:
$$ T_r = \frac{16}{r+1} \cdot \frac{1}{\frac{15}{r+1} \cdot ^{14}C_r} $$

$$ T_r = \frac{16}{15 \cdot ^{14}C_r} $$

Now, we calculate the product of these terms from $r=0$ to $12$:
$$ P = \prod_{r=0}^{12} T_r = \prod_{r=0}^{12} \frac{16}{15 \cdot ^{14}C_r} $$

Since the term $\frac{16}{15}$ is constant, and there are $13$ terms (from 0 to 12):
$$ P = \left( \frac{16}{15} \right)^{13} \cdot \frac{1}{^{14}C_0 \cdot ^{14}C_1 \cdots ^{14}C_{12}} $$

Comparing this with the given expression $\frac{\alpha^{13}}{^{14}C_0 \cdot ^{14}C_1 \cdots ^{14}C_{12}}$:
$$ \alpha^{13} = \left( \frac{16}{15} \right)^{13} \implies \alpha = \frac{16}{15} $$

We need to find the value of $30\alpha$:
$$ 30\alpha = 30 \times \frac{16}{15} $$

$$ 30\alpha = 2 \times 16 = 32 $$

Ans. 32