Sums of Squares and Cubes of First n Natural Numbers (JEE Advanced)

“sum of the first n natural numbers, sums of squares and cubes of the first n natural numbers”

Source: JEE Advanced 2026 Information Bulletin, Annexure-I. Closest JEE Main topic: Arithmetic and Geometric progressions, insertion of arithmetic, geometric means between two given numbers.

JEE Main focuses on AP and GP formulas, but Advanced expects you to handle sums involving higher powers of natural numbers. You need these formulas for problems combining series, mathematical induction, and integral approximations.

• The sum of first n natural numbers is n(n+1)/2, which you probably already know from AP formula
• Sum of squares formula n(n+1)(2n+1)/6 can be derived using telescoping with (k+1)³ - k³
• Sum of cubes equals the square of sum of first n numbers: [n(n+1)/2]², a beautiful identity
• These formulas are essential when converting Riemann sums to definite integrals in limits
• Problems often combine these: you might need Σk + Σk² + Σk³ in a single expression
• Mathematical induction proofs for these formulas are classic JEE Advanced questions

• Σk (k=1 to n) = n(n+1)/2
• Σk² (k=1 to n) = n(n+1)(2n+1)/6
• Σk³ (k=1 to n) = [n(n+1)/2]² = (Σk)²
• Σk⁴ (k=1 to n) = n(n+1)(2n+1)(3n²+3n-1)/30
• 1³ + 2³ + 3³ + ... + n³ = (1 + 2 + 3 + ... + n)²
• For large n: Σk²/n³ → 1/3 and Σk³/n⁴ → 1/4 (useful in Riemann sums)

• Confusing the sum of cubes formula with cube of sum: Σk³ = (Σk)², not Σ(k³) = (Σk)³
• Forgetting the factor of 6 in denominator of sum of squares formula and writing n(n+1)(2n+1)/2
• In limit problems, not extracting the correct power of n from the denominator before applying formulas
• Sign errors when using telescoping method: (k+1)³ - k³ = 3k² + 3k + 1, not 3k² + 3k - 1

Question. Evaluate: lim(n→∞) [1³ + 2³ + 3³ + ... + n³] / [n²(1² + 2² + ... + n²)]

Solved twice independently; both solutions agree.