In the 1967 MAA High School Mathematics Contest (Problem #39), contestants encountered an intriguing sequence of number sets:

"Given the sets of consecutive integers $\{1\}$, $\{2, 3\}$, $\{4, 5, 6\}$, $\{7, 8, 9, 10\}, \dots$, where each set contains one more element than the preceding one. What is the sum of the elements in the 20th set?"

Let's analyze the pattern:

• Set 1: $\{1\}$, length = 1, sum = 1
• Set 2: $\{2, 3\}$, length = 2, sum = 5
• Set 3: $\{4, 5, 6\}$, length = 3, sum = 15
• Set 4: $\{7, 8, 9, 10\}$, length = 4, sum = 34

Notice where each set begins! Before Set $n$, all previous sets have used up a number of integers equal to the $(n-1)$-th triangular number: $$T_{n-1} = 1 + 2 + 3 + \dots + (n - 1) = \frac{(n-1)n}{2} = \frac{n^2 - n}{2}$$ Therefore, the first number in Set $n$ is: $$\text{first}(n) = T_{n-1} + 1 = \frac{n^2 - n + 2}{2}$$ Since Set $n$ contains $n$ consecutive numbers, its last number is $\text{first} + (n - 1) = \frac{n^2 + n}{2}$.

Using Gauss's formula for the sum of an arithmetic progression ($\text{sum} = n \times \text{average}$): $$S(n) = n \cdot \left(\frac{\text{first} + \text{last}}{2}\right) = n \cdot \left(\frac{n^2 + 1}{2}\right) = \frac{n(n^2 + 1)}{2}$$ For the 20th set ($n = 20$): $$S(20) = \frac{20(20^2 + 1)}{2} = 10 \cdot 401 = 4010$$ In this lesson, we write nested loops to generate these sets, sum them up, and confirm our algebraic formula!

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• Triangular Numbers: The numbers $1, 3, 6, 10, 15, 21, \dots$ are called triangular numbers because that many dots can be arranged into an equilateral triangle. They appear constantly in combinatorics and number patterns.
• Nicomachus's Theorem: If you partition consecutive odd integers instead: $\{1\}, \{3, 5\}, \{7, 9, 11\}, \dots$, the sum of the $n$-th set is $n^3$! (e.g. $1 = 1^3$, $3+5 = 8 = 2^3$, $7+9+11 = 27 = 3^3$).