JEE Main 2026 — Mathematics Algebra

The given series is $S = 1^3 - 2^3 + 3^3 - 4^3 + \ldots + 15^3$S=13−23+33−43+…+153.

This can be rewritten as the sum of all cubes up to $15$15 minus twice the sum of cubes of even numbers up to $14$14:

$$S = (1^3 + 2^3 + 3^3 + \ldots + 15^3) - 2(2^3 + 4^3 + 6^3 + \ldots + 14^3)$$S=(13+23+33+…+153)−2(23+43+63+…+143)

$$S = \sum_{k=1}^{15} k^3 - 2 \times 2^3 (1^3 + 2^3 + 3^3 + \ldots + 7^3)$$S=k=1∑15​k3−2×23(13+23+33+…+73)

$$S = \sum_{k=1}^{15} k^3 - 16 \sum_{k=1}^{7} k^3$$S=k=1∑15​k3−16k=1∑7​k3

Using the formula for the sum of cubes $\displaystyle \sum_{k=1}^{n} k^3 = \left( \dfrac{n(n+1)}{2} \right)^2$k=1∑n​k3=(2n(n+1)​)2:

$$\sum_{k=1}^{15} k^3 = \left( \dfrac{15 \times 16}{2} \right)^2 = (120)^2 = 14400$$k=1∑15​k3=(215×16​)2=(120)2=14400

$$\sum_{k=1}^{7} k^3 = \left( \dfrac{7 \times 8}{2} \right)^2 = (28)^2 = 784$$k=1∑7​k3=(27×8​)2=(28)2=784

Substituting these values back into the expression for $S$S:

$$S = 14400 - 16 \times 784$$S=14400−16×784

$$S = 14400 - 12544 = 1856$$S=14400−12544=1856

Answer: $1856$1856