Understanding Triangular Numbers

A triangular number is a number obtained by adding all positive integers up to a given positive integer. It represents the number of dots required to form a triangle with sides of equal length, where each row contains one more dot than the row above it, starting with one dot in the first row.

The formula for the n-th triangular number, denoted as \(T_n\), is given by:

\[ T_n = \frac{n(n+1)}{2} \]

We need to find the first five triangular numbers.

Calculating the First Five Triangular Numbers

Using the formula \(T_n = \frac{n(n+1)}{2}\), we can find the first five triangular numbers:

• For \(n=1\): \(T_1 = \frac{1(1+1)}{2} = \frac{1 \times 2}{2} = \frac{2}{2} = 1\)
• For \(n=2\): \(T_2 = \frac{2(2+1)}{2} = \frac{2 \times 3}{2} = \frac{6}{2} = 3\)
• For \(n=3\): \(T_3 = \frac{3(3+1)}{2} = \frac{3 \times 4}{2} = \frac{12}{2} = 6\)
• For \(n=4\): \(T_4 = \frac{4(4+1)}{2} = \frac{4 \times 5}{2} = \frac{20}{2} = 10\)
• For \(n=5\): \(T_5 = \frac{5(5+1)}{2} = \frac{5 \times 6}{2} = \frac{30}{2} = 15\)

So, the first five triangular numbers are 1, 3, 6, 10, and 15.

Calculating the Mean of the First Five Triangular Numbers

The mean (or average) of a set of numbers is calculated by summing all the numbers in the set and then dividing by the count of numbers in the set.

Mean = \(\frac{\text{Sum of numbers}}{\text{Count of numbers}}\)

In this case, the numbers are the first five triangular numbers: 1, 3, 6, 10, 15.

Sum of the first five triangular numbers = \(1 + 3 + 6 + 10 + 15 = 35\)

Count of numbers = 5

Now, we calculate the mean:

Mean = \(\frac{35}{5} = 7\)

The mean of the first five triangular numbers is 7.

The sum is \(1+3+6+10+15 = 35\). The count is 5. The mean is \(35 \div 5 = 7\).

Revision Table: Key Concepts

Additional Information: Properties of Triangular Numbers

Triangular numbers have several interesting properties:

• The sum of two consecutive triangular numbers is a perfect square. For example, \(T_3 + T_4 = 6 + 10 = 16 = 4^2\). \(T_4 + T_5 = 10 + 15 = 25 = 5^2\). In general, \(T_n + T_{n+1} = n^2\). No, this property is incorrect. The correct property is \(T_n + T_{n-1} = n^2\). Let's verify: \(T_n = \frac{n(n+1)}{2}\) and \(T_{n-1} = \frac{(n-1)n}{2}\). \(T_n + T_{n-1} = \frac{n(n+1)}{2} + \frac{n(n-1)}{2} = \frac{n^2+n+n^2-n}{2} = \frac{2n^2}{2} = n^2\). Yes, that's correct. So the sum of the n-th and (n-1)-th triangular numbers is \(n^2\). For example, \(T_5 + T_4 = 15 + 10 = 25 = 5^2\). \(T_4 + T_3 = 10 + 6 = 16 = 4^2\).
• Every perfect number is a triangular number (except for 6). A perfect number is a positive integer that is equal to the sum of its proper positive divisors (excluding the number itself). For example, 6 is a perfect number (\(1+2+3=6\)) and \(T_3 = 6\). The next perfect number is 28, and \(T_7 = \frac{7(8)}{2} = 28\).
• The sum of the first n triangular numbers is given by the formula: \(\sum_{k=1}^{n} T_k = \frac{n(n+1)(n+2)}{6}\). This is the n-th tetrahedral number. For the first 5 triangular numbers: \(\frac{5(5+1)(5+2)}{6} = \frac{5 \times 6 \times 7}{6} = 5 \times 7 = 35\), which matches our sum.