JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Let [·] denote the greatest integer function. We have f(x) = _ n 1 n^3 _ k=1 ^n [ k^2 3^x ]. For large n: _ k=1 ^n [ k^2 3^x ] _ k=1 ^n k^2 3^x = 1 3^x n(n+1)(2n+1) 6 n^3 3 3^x Therefore: f(x) = _ n 1 n^3 n^3 3 3^x = 1 3^ x+1 Computing the infinite series: _ j=1 ^ f(j) = _ j=1 ^ 1 3^ j+1 = 1 3^2 + 1 3^3 + 1 3^4 + = 1 9 (1 + 1 3 + 1 9 + ) = 1 9 1 1 - 1/3 = 1 9 3 2 = 1 6 Therefore: 12 _ j=1 ^ f(j) = 12 1 6 = 2