JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

The given series is S = 1^3 - 2^3 + 3^3 - 4^3 + + 15^3. This can be rewritten as the sum of all cubes up to 15 minus twice the sum of cubes of even numbers up to 14: S = (1^3 + 2^3 + 3^3 + + 15^3) - 2(2^3 + 4^3 + 6^3 + + 14^3) S = _ k=1 ^ 15 k^3 - 2 2^3 (1^3 + 2^3 + 3^3 + + 7^3) S = _ k=1 ^ 15 k^3 - 16 _ k=1 ^ 7 k^3 Using the formula for the sum of cubes _ k=1 ^ n k^3 = ( n(n+1) 2 )^2: _ k=1 ^ 15 k^3 = ( 15 16 2 )^2 = (120)^2 = 14400 _ k=1 ^ 7 k^3 = ( 7 8 2 )^2 = (28)^2 = 784 Substituting these values back into the expression for S: S = 14400 - 16 784 S = 14400 - 12544 = 1856 Answer: 1856