JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Let S = _ n=1 ^ k f(n), where f(n) = vmatrix n & -1 & -5 \\ -2n^2 & 3(2k+1) & 2k+1 \\ -3n^3 & 3k(2k+1) & 3k(k+2)+1 vmatrix Since only the first column depends on n, the summation can be taken inside that column: S = vmatrix _ n=1 ^ k n & -1 & -5 \\ _ n=1 ^ k (-2n^2) & 3(2k+1) & 2k+1 \\ _ n=1 ^ k (-3n^3) & 3k(2k+1) & 3k(k+2)+1 vmatrix Using standard summation formulas: _ n=1 ^ k n = k(k+1) 2 _ n=1 ^ k (-2n^2) = - k(k+1)(2k+1) 3 _ n=1 ^ k (-3n^3) = - 3k^2(k+1)^2 4 Substituting: S = vmatrix k(k+1) 2 & -1 & -5 \\ - k(k+1)(2k+1) 3 & 3(2k+1) & 2k+1 \\ - 3k^2(k+1)^2 4 & 3k(2k+1) & 3k^2+6k+1 vmatrix Taking k(k+1) 12 common from C_1: S = k(k+1) 12 vmatrix 6 & -1 & -5 \\ -4(2k+1) & 3(2k+1) & 2k+1 \\ -9k(k+1) & 3k(2k+1) & 3k^2+6k+1 vmatrix Applying C_1 C_1 + C_2 + C_3: Row 1: 6 - 1 - 5 = 0 Row 2: -8k - 4 + 6k + 3 + 2k + 1 = 0 Row 3: -9k^2 - 9k + 6k^2 + 3k + 3k^2 + 6k + 1 = 1 S = k(k+1) 12 vmatrix 0 & -1 & -5 \\ 0 & 3(2k+1) & 2k+1 \\ 1 & 3k(2k+1) & 3k^2+6k+1 vmatrix Expanding along C_1: S = k(k+1) 12 1 vmatrix -1 & -5 \\ 3(2k+1) & 2k+1 vmatrix S = k(k+1) 12 [(-1)(2k+1) - (-5)(3(2k+1)) ] S = k(k+1) 12 [-(2k+1) + 15(2k+1) ] S = k(k+1) 12 14(2k+1) = 7 k(k+1)(2k+1) 6 Note that k(k+1)(2k+1) 6 = _ n=1 ^ k n^2. Given S = 98: 7 k(k+1)(2k+1) 6 = 98 k(k+1)(2k+1) 6 = 14 Checking integer values of k: k = 1: sum = 1 k = 2: sum = 5 k = 3: sum = 14 Hence, k = 3, and the correct option is (1)\ 3.