JEE Main 2026 — Mathematics Algebra

The general term of the product is $T_r = \frac{1}{^{15}C_{r-1}} + \frac{1}{^{15}C_r}$Tr​=15Cr−1​1​+15Cr​1​ for $r = 1, 2, \dots, 13$r=1,2,…,13.

Using the formula $^{n}C_r = \frac{n}{r} \cdot ^{n-1}C_{r-1}$nCr​=rn​⋅n−1Cr−1​, we have $^{15}C_r = \frac{15}{r} \cdot ^{14}C_{r-1}$15Cr​=r15​⋅14Cr−1​ and $^{15}C_{r-1} = \frac{15}{15-(r-1)} \cdot ^{14}C_{r-1} = \frac{15}{16-r} \cdot ^{14}C_{r-1}$15Cr−1​=15−(r−1)15​⋅14Cr−1​=16−r15​⋅14Cr−1​.

Substituting these into $T_r$Tr​:

$T_r = \frac{16-r}{15 \cdot ^{14}C_{r-1}} + \frac{r}{15 \cdot ^{14}C_{r-1}} = \frac{16-r+r}{15 \cdot ^{14}C_{r-1}} = \frac{16}{15 \cdot ^{14}C_{r-1}}$Tr​=15⋅14Cr−1​16−r​+15⋅14Cr−1​r​=15⋅14Cr−1​16−r+r​=15⋅14Cr−1​16​.

The product is $\displaystyle \prod_{r=1}^{13} T_r= \prod_{r=1}^{13} \frac{16}{15\cdot {^{14}C_{r-1}}}= \frac{16^{13}}{15^{13} \cdot \prod_{r=1}^{13}{^{14}C_{r-1}}}$r=1∏13​Tr​=r=1∏13​15⋅14Cr−1​16​=1513⋅∏r=113​14Cr−1​1613​.

The denominator of the product is $15^{13} \cdot (^{14}C_0 \cdot ^{14}C_1 \dots ^{14}C_{12})$1513⋅(14C0​⋅14C1​…14C12​).

Comparing this with the given expression $\frac{\alpha^{13}}{^{14}C_0 \cdot ^{14}C_1 \dots ^{14}C_{12}}$14C0​⋅14C1​…14C12​α13​, we get:

$\alpha^{13} = \frac{16^{13}}{15^{13}} \implies \alpha = \frac{16}{15}$α13=15131613​⟹α=1516​.

We need to find the value of $30\alpha$30α:

$30\alpha = 30 \times \frac{16}{15} = 2 \times 16 = 32$30α=30×1516​=2×16=32.