Sum of Squares — First 35 Natural Numbers Find the value of 1^2 + 2^2 + 3^2 + … + 35^2 using the standard formula.

Introduction / Context:
Summing squares is a frequent task in quantitative aptitude and statistics. Rather than computing each square, rely on the closed-form formula for the sum of the first n squares to obtain the result quickly and accurately.

Given Data / Assumptions:

• We need 1^2 + 2^2 + … + 35^2.
• n = 35 is a positive integer.

Concept / Approach:
The standard formula is S = n(n + 1)(2n + 1)/6. Substitute n = 35 and compute carefully by simplifying before multiplying to reduce arithmetic effort and minimize errors.

Step-by-Step Solution:
Identify n = 35 → n + 1 = 36, 2n + 1 = 71.Apply S = n(n + 1)(2n + 1)/6 = 35 * 36 * 71 / 6.Simplify 36 / 6 = 6 → S = 35 * 6 * 71.Compute 35 * 6 = 210; then 210 * 71 = 210 * (70 + 1) = 14,700 + 210 = 14,910.

Verification / Alternative check:
Estimate: average square near (35/2)^2 ≈ 17.5^2 ≈ 306; 35 terms × 306 ≈ 10,710 (a rough underestimate). Precise formula gives 14,910—consistent with exact calculation, not the rough estimate.

Why Other Options Are Wrong:
15510, 14510, 16510, and 15210 arise from slips such as not reducing by 6 first, mis-multiplying 210 * 71, or transposing digits.

Common Pitfalls:
Multiplying 35 * 36 * 71 first (risking large-number errors) instead of simplifying by 6; arithmetic mistakes when distributing 210 across 71.

Final Answer:
14910