Last modified on December 26th, 2023

chapter outline

 

Triangular Number

A triangular number is a number that can be expressed as the sum of the first n consecutive positive integers starting from 1. The numbers form a sequence:

1, 3, 6, 10, 15, 21…. which continues till infinity.

In the triangular number sequence:

  • The first number is 1
  • The second number is (1 + 2) = 3
  • The third triangular number is (1 + 2 + 3) = 6, and so on

Let us imagine laying out objects in the shape of an equilateral triangle. If we place one object in the first row, two objects in the second row, three in the third, and so on, the total number of objects will form an equilateral triangle. Hence, the name ‘triangular number’ is given to each number of this sequence. 

If a dot represents each object, the equilateral triangles will look as shown:

Triangular Number

Thus, the nth triangular number (Tn) is represented by n dots on each side of the triangle, and the total number of dots is obtained by the sum of n consecutive positive numbers from 1 to n.

For example, to find the 50th triangular number, we would have to add 50 successive positive numbers. 

However, adding 50 such numbers will take time and effort. Instead, we can use a formula to simplify our calculation to find any given triangular number.

Formula 

The formula to find the nth triangular number is: 

${T_{n}=\sum ^{n}_{i=1}i=\dfrac{n\left( n+1\right) }{2}}$

Derivation

As we know, the sum of the first n natural numbers can be expressed as an arithmetic series. The series starts with 1 and increments by 1 at each step, so it can be represented as follows:

1 + 2 + 3 + 4 + … + n

Summation (Σ) of the above sequence gives

${\sum ^{n}_{i=1}i}$

Now, we know the sum of n terms in an arithmetic series is:

${\sum ^{n}_{i=1}i=\dfrac{n\left( n+1\right) }{2}}$

Thus, for the nth triangular number, 

${T_{n}=\sum ^{n}_{i=1}i=\dfrac{n\left( n+1\right) }{2}}$, here ${\dfrac{n\left( n+1\right) }{2}}$ is the binomial coefficient, which represents the number of distinct pairs that can be choosen from n + 1 objects.

Triangular Numbers Formula

Thus, the sum of n natural numbers forms a triangular number. We can also observe that each term (from 1 to n) in Tn has a common difference of one. 

Now, let us find the 50th triangular number using the above formula.

As we know, ${T_{n}=1+2+3+\ldots +n=\dfrac{n}{2}\left( n+1\right)}$

Here, n = 50, ${T_{50}=\dfrac{50}{2}\left( 50+1\right)}$ = 25 x 51 = 1275

Thus, the 50th triangular number is 1275.

Patterns and Relationships

To Rectangular Numbers

If we arrange each triangular number in the shape of a right-angled triangle and then double the number of dots on its hypotenuse, it forms a rectangular shape.

Triangular Number Pattern

To Square Numbers

As we know, the sum of two consecutive natural numbers always forms a square number.

T1 + T2 = 1 + 3 = 4 = 22

T2 + T3 = 3 + 6 = 9 = 32

Similarly, we can use the formula to find the sum of the triangular numbers Tn and Tn + 1;

Tn  + Tn + 1 = ${\dfrac{n}{2}\left( n+2\right) +\dfrac{\left( n+1\right) \left( n+2\right) }{2}}$

= ${\dfrac{\left( n+1\right) \left( 2n+2\right) }{2}}$

= ${\left( n+1\right) ^{2}}$

Thus, the sum of two consecutive triangular numbers always gives the square numbers in the results.

The following table lists a few square numbers.

Triangular numberSum of two consecutive Triangular numbersSquare number
11${1^{2}}$
31 + 3 =  4${2^{2}}$
63 + 6 = 9${3^{2}}$
106 + 10 = 16${4^{2}}$
1510 + 15 = 25${5^{2}}$
2115 + 21 = 36${6^{2}}$
2821 + 28 = 49${7^{2}}$
3628 + 36 = 64${8^{2}}$
4536 + 45 = 81${9^{2}}$
5545 + 55 = 100${10^{2}}$
6655 + 66 = 121${11^{2}}$
7866 + 78 = 144${12^{2}}$

Solved Examples

Are 98 and 72 triangular numbers?

Solution:

As we know, triangular numbers are obtained by adding consecutive positive numbers.
The sequence is: 
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105 …
Thus, 98 and 72 are not triangular numbers.

Find the following triangular numbers.

Solution:

a.1000th
b.35th
c.30th
d.100th
As we know, ${T_{n}=1+2+3+\ldots +n=\dfrac{n}{2}\left( n+1\right)}$
a. Here n = 1000
Thus, the 1000th triangular number is T1000 = ${\dfrac{1000}{2}\left( 1000+1\right)}$
= 500 x 1001 = 500500
b. Here n = 35
Thus, the 35th triangular number is T35 = ${\dfrac{35}{2}\left( 35+1\right)}$
= 35 x 18 = 630
c. Here n = 30
Thus, the 30th triangular number is T30 = ${\dfrac{30}{2}\left( 30+1\right)}$
= 15 x 31 = 465
d. Here n = 100
Thus, the 100th triangular numbers is T100 = ${\dfrac{100}{2}\left( 100+1\right)}$
= 50 x 101 = 5050

Last modified on December 26th, 2023

Comments are closed.