Sum Of Cubes

The sum of the first n cubes equals the square of the sum of the first n natural numbers, a striking special-series identity exploited in JEE Advanced telescoping arguments. Since 1 + 2 + ... + n = n(n+1)/2, the cube sum equals [n(n+1)/2]^2. This can be proved by induction or by telescoping the identity k^4 - (k-1)^4. Option n(n+1)(2n+1)/6 is the sum of squares, not cubes. Option n(n+1)/2 is merely the linear sum, missing the squaring. Option [n(n+1)]^2 omits the division by two and overcounts by a factor of four. Hence the closed form is [n(n+1)/2]^2. This identity, sometimes called Nicomachus's theorem, also shows that every sum of consecutive cubes from one is automatically a perfect square, a fact JEE Advanced occasionally hides inside number-theoretic dressing to test recognition. Plausibility check: for n = 3 the direct sum is 1 + 8 + 27 = 36, and [3·4/2]^2 = 6^2 = 36, matching exactly and confirming the elegant square-of-triangular-number pattern.