JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Let S = 2^3 3 \ ^ 12 C_ 2 + 2^5 5 \ ^ 12 C_ 4 + 2^7 7 \ ^ 12 C_ 6 + + 2^ 13 13 \ ^ 12 C_ 12 Using the property 1 r+1 \ ^ n C_ r = 1 n+1 \ ^ n+1 C_ r+1 , we can rewrite the terms: S = 2^3 13 \ ^ 13 C_ 3 + 2^5 13 \ ^ 13 C_ 5 + 2^7 13 \ ^ 13 C_ 7 + + 2^ 13 13 \ ^ 13 C_ 13 13S = 2^3 \ ^ 13 C_ 3 + 2^5 \ ^ 13 C_ 5 + 2^7 \ ^ 13 C_ 7 + + 2^ 13 \ ^ 13 C_ 13 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 + + \ ^ 13 C_ 13 x^ 13 (1-x)^ 13 = \ ^ 13 C_ 0 - \ ^ 13 C_ 1 x + \ ^ 13 C_ 2 x^2 - \ ^ 13 C_ 3 x^3 + - \ ^ 13 C_ 13 x^ 13 Subtracting the second equation from the first gives: (1+x)^ 13 - (1-x)^ 13 = 2 ( \ ^ 13 C_ 1 x + \ ^ 13 C_ 3 x^3 + \ ^ 13 C_ 5 x^5 + + \ ^ 13 C_ 13 x^ 13 ) Substituting x = 2: (1+2)^ 13 - (1-2)^ 13 = 2 ( \ ^ 13 C_ 1 (2) + \ ^ 13 C_ 3 (2^3) + \ ^ 13 C_ 5 (2^5) + + \ ^ 13 C_ 13 (2^ 13 ) ) 3^ 13 - (-1)^ 13 = 2 ( 26 + 13S ) 3^ 13 + 1 = 52 + 26S 26S = 3^ 13 + 1 - 52 26S = 3^ 13 - 51 Comparing this with the given expression 26S = 3^ 13 - , we get: = 51 Answer: 51