Series summation — cubes of natural numbers Find the exact sum of the cubes of the first 15 natural numbers (i.e., 1^3 + 2^3 + … + 15^3).

Introduction / Context: Summations of powers appear frequently in number system and algebra questions. A classic identity states that the sum of the first n cubes equals the square of the nth triangular number. Applying the identity correctly saves time and avoids long arithmetic.

Given Data / Assumptions:

• We need S = 1^3 + 2^3 + … + 15^3.
• Use the standard closed-form: sum of first n cubes.
• Basic integer arithmetic is sufficient; no calculators required if steps are organized.

Concept / Approach: The identity is: sum_{k=1 to n} k^3 = (n*(n+1)/2)^2. That means we first compute the triangular sum T = 1 + 2 + … + n = n*(n+1)/2, and then square T to obtain the sum of cubes. This both reduces effort and limits error risk.

Step-by-Step Solution:Let n = 15.Compute T = n*(n+1)/2 = 15*16/2 = 120.Square T: S = T^2 = 120^2 = 14400.Therefore, the required sum is 14400.

Verification / Alternative check: Spot-check by partial sums (e.g., up to 5: 1+8+27+64+125 = 225, and (5*6/2)^2 = 15^2 = 225) to reinforce confidence in the identity, then rely on the same logic for n = 15.

Why Other Options Are Wrong:15400, 16800, 13300 result from squaring an incorrect triangular sum or arithmetic slips in 120^2.

Common Pitfalls: Forgetting to square the triangular sum; computing 15*16 incorrectly; mixing the square of the sum with the sum of squares by mistake. Always remember: sum of cubes = (sum of first n integers)^2.

Final Answer: 14400