Triangular number: Definition

0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, 496, 528, 561, 595, 630, 666, 703, 741, 780, 820, 861, 903, 946, 990, 1035, 1081, 1128, 1176, 1225, 1275, 1326, 1378, 1431

T(1) = 1

T(2) = 1 + 2 = 3

T(3) = 1 + 2 + 3 = 6

T(4) = 1 + 2 + 3 + 4 = 10

T(5) = 1 + 2 + 3 + 4 + 5 = 15

Then it is clear that T(n) is the sum of the first n positive integers. So the n-th triangular number can be obtained as **T(n) = n*(n+1)/2**, where **n** is any natural number

The sum of any two consecutive triangular numbers is alway a perfect square:

E.g. 1+3=4, 3+6=9, 6+10=16 and so on.

Generally, T(n-1) and T(n) are two consecutive terms. We have,

**T(n-1) + T(n) = (n-1)n/2 + n(n+1)/2 = n^2**

**>> Let T be a triangular number, then 8*T + 1 is a perfect square**

For example,

8*1 + 1 = 9 = 3^2

8*3 + 1 = 25 = 5^2

8*6 + 1 = 49 = 7^2

8*10 + 1 = 81 = 9^2

and so on.

We can write, **8*T(n) + 1 = (2n + 1)^2**

T(n) = n*(n+1)/2

8*T(n) + 1 = 8*n*(n+1)/2 + 1 = 4*n*(n+1) + 1 = 4n^2 + 4n + 1 = (2n + 1)^2

**>> Sum of the squares of two consecutive triangular numbers**

(T(1))^2 + (T(2))^2 = 1^2 + 3^2 = 10 is a triangular number

(T(2))^2 + (T(3))^2 = 3^2 + 6^2 = 45 is a triangular number

Generally, **(T(n-1))^2 + (T(n))^2 **

The sum of the squares is:

(n-1)^2 * n^2/4 + n^2 * (n+1)^2/4

n^2/4 * { (n-1)^2 + (n+1)^2 }

n^2/4 * (n^2 – 2n + 1 + n^2 + 2n + 1)

n^2/4 * (2n^2 + 2)

n^2 * (n^2 + 1)/2, which is **T(n^2)**

So the sum of the squares of two consecutive triangular numbers is a triangular number.

**>> If T is a triangular number, so is 9*T + 1**

For example, 1, 3, 6 are triangular numbers

9*1 + 1 = 10 (triangular number)

9*3 + 1 = 28 (triangular number)

9*6 + 1 = 55 (triangular number)

an so on

Generally, T(n) = n*(n + 1)/2

9*T(n) + 1 = 9*n*(n + 1)/2 + 1

(9n^2 + 9n + 2)/2

(3n + 1)(3n + 2)/2, which is **T(3n+1)**

**>> If T is a triangular number, so is 25*T + 3**

For example, 1, 3, 6 are triangular numbers

25*1 + 3 = 28 (triangular number)

25*3 + 3 = 78 (triangular number)

25*6 + 3 = 153 (triangular number)

Generally, T(n) = n*(n + 1)/2

25*T(n) + 3 = 25*n*(n + 1)/2 + 3

(25n^2 + 25n)/2 + 3

(25n^2 + 25n + 6)/2

(5n + 2)(5n + 3)/2, which is **T(5n+2)**

**>> If x is a Triangular number, so is ax+b**

We need to find all pairs (a,b) of positive integers which satisfy:

**If x is a Triangular number, then so is ax+b**

We saw above that **9*T + 1** and **25*T + 3** are triangular numbers.

The pairs (9, 1) and (25, 3) satisfy this condition.

The integers pairs (a, b) for which it holds are the following:

**a = an odd square, b = (a-1)/8**

So the pairs (a, b) are: (9,1), (25,3), (81,10), (121,15), …

Notice that **b** defined as **(a-1)/8** is itself a triangular number

**Questions:**

**>> T(1) = 1, T(7) = 28 and T(10) = 55 is an example of 3 triangular numbers in arithmetic progression. Find others.**

**>> Can you find 3 triangular numbers in geometric progression?**