**Fun with Triangle and Square Numbers**

**Formulas**

Triangle
Numbers:

Tn
= COMB(n +1, 2) = (n^2 + n)/2

Square
Numbers:

Sn
= n^2

where
n is an integer, n ≥ 1

**Mathematics between Triangle and Square Numbers**

**2*Tn**

2*Tn
= 2 * (n^2 + n)/2 = n^2 + n = Sn + n

Note
that 2*Tn is a rectangle number, which consists of a square with a row or
column attached to it. 2*Tn is an
integer.

**3*Tn**

3*Tn
= 3 * (n^2 + n)/2

We
can show that 3*Tn is an integer by showing that 3*Tn is an integer for both
even and odd n.

First
assume that n is even, let n = 2*k and k is a positive integer:

3
* Tn

=
3/2 * ((2*k)^2 + 2*k)

=
3/2 * (4*k^2 + 2*k)

=
3 * (2*k^2 + k)

The
result is an integer.

Next,
let n be an odd integer of the form of n = 2*k + 1. Then

3
* Tn

=
3/2 * ((2*k + 1)^2 + (2*k + 1))

=
3/2 * (4*k^2 + 4*k + 1 + 2*k + 1)

=
3/2 * (4*k^2 + 6*k + 2)

=
3 * (2*k^2 + 3*k + 1)

The
result is an integer.

**Tn^2**

Tn^2
= (n^2 + n)/2)^2

Expanding
Tn^2 to get:

=
(n^4 + 2*n^3 + n^2)/4

Tn^2
in terms of Sn and n:

Tn^2

=
(Sn^2 +2*n^3 + Sn)/4

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

=
(Sn^2 + Sn*(2*n + 1))/4

Clearly
Tn^2 is an integer since Tn is an integer, and multiplying two integers
generates another integer.

**Square Numbers in terms of Triangle Numbers**

With
Sn = n^2 and Tn = (n^2 + n)/2

Tn
= (n^2 + n)/2

Tn
= (Sn + n)/2

2
* Tn = Sn + n

2
* Tn – n = Sn

Eddie

This
blog is property of Edward Shore, 2018.
(Happy New Year!)

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