JEE Main 2026 — Mathematics Algebra

The expansion of $(1 + \alpha x)^{26}$(1+αx)26 has $27$27 terms. The middle term is $T_{14}$T14​.

The coefficient of the middle term is $^{26}C_{13} \alpha^{13}$26C13​α13.

The expansion of $(1 - \alpha x)^{28}$(1−αx)28 has $29$29 terms. The middle term is $T_{15}$T15​.

The coefficient of the middle term is $^{28}C_{14} (-\alpha)^{14} = ^{28}C_{14} \alpha^{14}$28C14​(−α)14=28C14​α14.

Equating the two coefficients:

$$^{26}C_{13} \alpha^{13} = ^{28}C_{14} \alpha^{14}$$26C13​α13=28C14​α14

Since $\alpha \neq 0$α=0, dividing by $\alpha^{13}$α13 gives:

$$\alpha = \dfrac{^{26}C_{13}}{^{28}C_{14}}$$α=28C14​26C13​​

Expanding the combinations using factorials:

$$^{28}C_{14} = \dfrac{28!}{14! 14!} = \dfrac{28 \times 27 \times 26!}{14 \times 13! \times 14 \times 13!} = \dfrac{28 \times 27}{14 \times 14} \times \dfrac{26!}{13! 13!}$$28C14​=14!14!28!​=14×13!×14×13!28×27×26!​=14×1428×27​×13!13!26!​

$$^{28}C_{14} = \dfrac{27}{7} \times ^{26}C_{13}$$28C14​=727​×26C13​

Substituting this back:

$$\alpha = \dfrac{^{26}C_{13}}{\dfrac{27}{7} \times ^{26}C_{13}} = \dfrac{7}{27}$$α=727​×26C13​26C13​​=277​

Answer: $\dfrac{7}{27}$277​