JEE Main 2026 — 6 April, Morning Shift — Question 25

We need to compute $S = 1^3 - 2^3 + 3^3 - 4^3 + \cdots + 15^3$. Group the terms as $(1^3 - 2^3) + (3^3 - 4^3) + \cdots + (13^3 - 14^3) + 15^3$. For any integer $n$, $n^3 - (n+1)^3 = -3n^2 - 3n - 1$. Here $n$ takes odd values: 1, 3, 5, ..., 13 (7 terms). Sum of the 7 pairs: $\sum_{k=0}^{6} [ -3(2k+1)^2 - 3(2k+1) - 1 ]$. Compute: $-3\sum (4k^2+4k+1) -3\sum (2k+1) - 7$. $\sum_{k=0}^{6} k^2 = 91$, $\sum_{k=0}^{6} k = 21$. So $-3(4\cdot91 + 4\cdot21 + 7) -3(2\cdot21+7) -7 = -3(364+84+7) -3(42+7) -7 = -3\cdot455 -3\cdot49 -7 = -1365 -147 -7 = -1519$. Add the last term $15^3 = 3375$: $S = 3375 - 1519 = 1856$. Thus the value is $1856.$