JEE Main 2026 — Mathematics Algebra

Let $S = \dfrac{2^3}{3} \ ^{12}C_{2} + \dfrac{2^5}{5} \ ^{12}C_{4} + \dfrac{2^7}{7} \ ^{12}C_{6} + \ldots + \dfrac{2^{13}}{13} \ ^{12}C_{12}$S=323​ 12C2​+525​ 12C4​+727​ 12C6​+…+13213​ 12C12​

Using the property $\dfrac{1}{r+1} \ ^{n}C_{r} = \dfrac{1}{n+1} \ ^{n+1}C_{r+1}$r+11​ nCr​=n+11​ n+1Cr+1​, we can rewrite the terms:

$$S = \dfrac{2^3}{13} \ ^{13}C_{3} + \dfrac{2^5}{13} \ ^{13}C_{5} + \dfrac{2^7}{13} \ ^{13}C_{7} + \ldots + \dfrac{2^{13}}{13} \ ^{13}C_{13}$$S=1323​ 13C3​+1325​ 13C5​+1327​ 13C7​+…+13213​ 13C13​

$$13S = 2^3 \ ^{13}C_{3} + 2^5 \ ^{13}C_{5} + 2^7 \ ^{13}C_{7} + \ldots + 2^{13} \ ^{13}C_{13}$$13S=23 13C3​+25 13C5​+27 13C7​+…+213 13C13​

We know the binomial expansions:

$$(1+x)^{13} = \ ^{13}C_{0} + \ ^{13}C_{1}x + \ ^{13}C_{2}x^2 + \ ^{13}C_{3}x^3 + \ldots + \ ^{13}C_{13}x^{13}$$(1+x)13= 13C0​+ 13C1​x+ 13C2​x2+ 13C3​x3+…+ 13C13​x13

$$(1-x)^{13} = \ ^{13}C_{0} - \ ^{13}C_{1}x + \ ^{13}C_{2}x^2 - \ ^{13}C_{3}x^3 + \ldots - \ ^{13}C_{13}x^{13}$$(1−x)13= 13C0​− 13C1​x+ 13C2​x2− 13C3​x3+…− 13C13​x13

Subtracting the second equation from the first gives:

$$(1+x)^{13} - (1-x)^{13} = 2 \left( \ ^{13}C_{1}x + \ ^{13}C_{3}x^3 + \ ^{13}C_{5}x^5 + \ldots + \ ^{13}C_{13}x^{13} \right)$$(1+x)13−(1−x)13=2( 13C1​x+ 13C3​x3+ 13C5​x5+…+ 13C13​x13)

Substituting $x = 2$x=2:

$$(1+2)^{13} - (1-2)^{13} = 2 \left( \ ^{13}C_{1}(2) + \ ^{13}C_{3}(2^3) + \ ^{13}C_{5}(2^5) + \ldots + \ ^{13}C_{13}(2^{13}) \right)$$(1+2)13−(1−2)13=2( 13C1​(2)+ 13C3​(23)+ 13C5​(25)+…+ 13C13​(213))

$$3^{13} - (-1)^{13} = 2 \left( 26 + 13S \right)$$313−(−1)13=2(26+13S)

$$3^{13} + 1 = 52 + 26S$$313+1=52+26S

$$26S = 3^{13} + 1 - 52$$26S=313+1−52

$$26S = 3^{13} - 51$$26S=313−51

Comparing this with the given expression $26S = 3^{13} - \alpha$26S=313−α, we get:

$$\alpha = 51$$α=51

Answer: $51$51