The square of the sum of three positive consecutive natural numbers exceeds the sum of their squares by 292. What is the largest of these three numbers?

Introduction / Context:
This problem combines algebra and properties of consecutive natural numbers. You are told that the square of the sum of three consecutive natural numbers is greater than the sum of their individual squares by 292. From this relationship you must determine the largest of the three numbers. This tests your ability to model consecutive numbers with variables and manipulate algebraic expressions involving squares and sums.

Given Data / Assumptions:

- The three numbers are positive consecutive natural numbers. - Let them be represented in terms of a central variable. - The square of their sum minus the sum of their squares equals 292. - We need to find the largest of these three numbers.

Concept / Approach:
For three consecutive integers, a convenient representation is n - 1, n and n + 1, where n is the middle number. This choice simplifies the algebra because the deviations from the middle are symmetric. We first compute the square of their sum, then the sum of their squares and form an equation using the given difference of 292. Solving the resulting quadratic equation in n produces the middle number, and hence the largest number n + 1.

Step-by-Step Solution:
Step 1: Let the three consecutive natural numbers be n - 1, n and n + 1, where n is a positive integer. Step 2: Compute their sum: (n - 1) + n + (n + 1) = 3n. Step 3: The square of their sum is (3n)^2 = 9n^2. Step 4: Compute the sum of their squares: (n - 1)^2 + n^2 + (n + 1)^2. Step 5: Expand each square: (n - 1)^2 = n^2 - 2n + 1, n^2 = n^2 and (n + 1)^2 = n^2 + 2n + 1. Step 6: Add these results: (n^2 - 2n + 1) + n^2 + (n^2 + 2n + 1) = 3n^2 + 2. Step 7: The difference between the square of the sum and the sum of the squares is 9n^2 - (3n^2 + 2) = 6n^2 - 2. Step 8: We are told this difference equals 292, so set 6n^2 - 2 = 292. Step 9: Solve for n: 6n^2 = 292 + 2 = 294, so n^2 = 294 / 6 = 49. Step 10: Thus n = 7, since n is positive. Step 11: The three numbers are n - 1 = 6, n = 7 and n + 1 = 8. The largest is 8.

Verification / Alternative check:
Verify the condition using the numbers 6, 7 and 8. Their sum is 6 + 7 + 8 = 21. The square of the sum is 21^2 = 441. The sum of their squares is 6^2 + 7^2 + 8^2 = 36 + 49 + 64 = 149. The difference is 441 - 149 = 292, matching the value given in the question. This confirms that 6, 7 and 8 are the correct numbers and 8 is the largest.

Why Other Options Are Wrong:
If the largest number were 5, then the numbers would be 3, 4 and 5 or 4, 5 and 6, and substituting these into the statement would not produce a difference of 292. Similar checks show that sets with largest numbers 6 or 7 cannot satisfy the required condition. Only the triple 6, 7 and 8 works, corresponding to largest number 8.

Common Pitfalls:
Some learners choose to represent the consecutive numbers as n, n + 1 and n + 2. Although this also works, it leads to slightly more complicated expressions. Others may square the sum incorrectly or forget to fully expand and simplify the sum of squares. Using the symmetric representation n - 1, n and n + 1 and working carefully through each algebraic step helps avoid these mistakes.

Final Answer:
The largest of the three positive consecutive natural numbers is 8, which corresponds to option D.