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Relation between Tetrahedral and Triangular numbers

Sum of squares vs. Triangular numbers

3D cube animation for sum of squares

Sum of integers squared from 1 to N is also called "Square Pyramid Numbers" because each layer of the balls
makes a square pattern.

It is expressed as Pyr_{n} = 1^{2} + 2^{2} + 3^{2} + ... + N^{2}

Instead ,if the cross section pattern is a triangle, then it makes the following number sequence.

1 = T_{1} = 1

1 + (1 + 2) = T_{1} + T_{2} = 4

1 + (1 + 2) + (1 + 2 + 3) = T_{1} + T_{2} + T_{3} = 10

1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) = T_{1} + T_{2} + T_{3} + T_{4} = 20

and so on.

This is the sum of the Triangular number,i.e. Tet_{n} = T_{1} + T_{2} + T_{3} + .... + T_{n}

This number is called Tetrahedral number.

It has been shown already that two consecutive triangular numbers makes a square number. Analogous rule exists for
square pyramid,i.e.,*a square pyramid is made up of two consecutive tetrahedral numbers.*

This relation between Square_Pyramid number and Tetrahedral number is easily understood by looking at the next two figures.

The figure below shows that One pyramid number is made up of two tetrahedral numbers, in this example

Pyr_{5} = Tet_{4} + Tet_{5}
and in general, Pyr_{n} = Tet_{n-1} + Tet_{n}

********

**To create this drawing : **
** Load sumsqr_1.lsp (load "sumsqr_1")**

Then from command line, type ** square_pyramid **

Spacing is chosen to be 2.0~2.5 . Closely packed when 1.5 .

The center plane is drawn by the command : **cut_plate**

Note* : This program requires**
red_ball.dwg ,yellow_ball.dwg ,green_ball.dwg ,cyan_ball.dwg
blue_ball.dwg ,magenta_ball.dwg ,white_ball.dwg**.

(Ref.1 p.45) shows a very interesting visual proof that the formula for n-th tetrahedral number is
** Tet _{n} = (1/6) n(n+1)(n+2) **

by using Triangular numbers.

The details of process can be seen in running animation for N=5 case.

**To create this drawing and animation: **
** Load sumsqr_3.lsp (load "sumsqr_3")
Then from command line, type sumsqr_3 **

Note* : This program requires

red_1_ring.dwg ,yellow_2_ring.dwg

green_3_ring.dwg ,cyan_4_ring.dwg

blue_5_ring.dwg ,magenta_6_ring.dwg

white_7_ring.dwg

*********************

In the figure;

Upper row --the first triangle : Tet

Upper row --the second triangle : rotate the first triangle 120 deg. clockwise using top apex as pivot .

Upper row --the third triangle : rotate the second triangle 120 deg. clockwise using top apex as pivot .

Lower row: adding all numbers in the circles at the same locations.Note that the total number of circles

are 5-th Triangulaer number(Tsub>5 = 1 + 2 + 3 + 4 + 5 )

So what this picture tells us is 3 x Tet

And in general, 3 x Tet

Then finally,

Now the Pyramid number can be expressed as follows.

1. Adding consecutive triangular number makes a square,or T

2. Triangular number : T

You can see the process in animation.

**To create this drawing and animation: **
** Load sumsqr_2.lsp (load "sumsqr_2")
Then from command line, type sumsqr_2 **

************

The example shown is for N=5 case. At the center line, draw a column of T

Then make N (= 5) copies of the T

Total number of columns is (2N + 1) (= 11).

Next, make three groups A, B, & C as shown.

Sums of numbers in 3 groups are done in the arrow direction in the figure.

Group A: Starting from top , 1 , 1+2+1, 1+2+3+2+1, 1+2+3+4+3+2+1,... Since each term is a sum of two consecutive

Triangular number, each term is a number squared. Sum of these S

Group B: 1+2

Group C: The same as group B. Sum is S

So we have (2N+1) copies of Trianglar number T

Therefore 3 S

or S

The following figures are the early and the final steps.

The details of process can be seen in running animation for N=5 case.

******** 3 identical blocks arranged parallel to x,y & z axes are assembled into one block.********

*****************sumsqr_1_start.dwg** **************
*****************sumsqr_1_final.dwg** **************

The volume of the final block is N(N+1){N+(1/2)} as shown above.

Since it is made up of three blocks, each of which is sum of squares 1 + 2^{2}+ 3^{2}+ ...+ N^{2},

3 (1 + 2^{2}+ 3^{2}+ ...+ N^{2}) = 3 S_{n} = N(N+1){N+(1/2)} ,or

S_{n} = (1/6)N(N+1)(2N+1)

**To create this drawing and animation: **
** Load sumsqr_1.lsp (load "sumsqr_1")
Then from command line, type sumsqr_1 **

Author's note:With some unknown reason, the program comes to a halt after 3 blocks are put together.

To continue the process, the following lines must be typed in the command line window.

explode !ss3 hit return key finish_up

Note* : This program requires

red_cube.dwg ,yellow_cube.dwg ,green_cube.dwg ,cyan_cube.dwg

blue_cube.dwg ,magenta_cube.dwg ,white_cube.dwg

- Conway,J.H., Guy,R.K.: The Book of Numbers. Springer-Verlag,New York, p 27, 1995.
- Nelson,R.B. : Proofs Without Words: Exercises in Visual Thinking. MAA, p.77, 1993.

Go to Fun_Math Content Table Sums of Integers and Series

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Last Updated Sept 9-th, 2006

Copyright 2006 Takaya Iwamoto All rights reserved. .