Aspire Faculty ID #17933 · Topic: JEE Main 2026 (21 January Evening Shift) · Just now
JEE Main 2026 (21 January Evening Shift)

If $\displaystyle \left(\frac{1}{^{15}C_0} + \frac{1}{^{15}C_1}\right)\left(\frac{1}{^{15}C_1} + \frac{1}{^{15}C_2}\right)\cdots\left(\frac{1}{^{15}C_{12}} + \frac{1}{^{15}C_{13}}\right) = \frac{\alpha^{13}}{^{14}C_0\cdot ^{14}C_1\cdots ^{14}C_{12}}$ then $30\alpha$ is equal to ______.

Solution

$\prod_{r=0}^{12} \left(\frac{1}{^{15}C_r} + \frac{1}{^{15}C_{r+1}}\right)$

$= \prod_{r=0}^{12} \frac{16}{(r+1)\cdot \frac{15!}{r!(15-r)!}}$

$= \prod_{r=0}^{12} \frac{16}{^{15}C_r \cdot \frac{r+1}{15}}$

$= \prod_{r=0}^{12} \frac{16}{^{14}C_r}$

$= \frac{\left(\frac{16}{15}\right)^{13}}{^{14}C_0\cdot ^{14}C_1\cdots ^{14}C_{12}}$

$\Rightarrow \alpha = \frac{16}{15}$

$\Rightarrow 30\alpha = 32$
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