Only available on StudyMode
  • Topic: Triangular number, Polygonal number, Square number
  • Pages : 8 (1208 words )
  • Download(s) : 63
  • Published : May 18, 2013
Open Document
Text Preview


Stellar numbers are sequence of numbers that follow a certain pattern, when we plot it into a diagram it will create a geometrical star-like shape. Each stellar number has its own vertices and number of dots, the formula will be different for every vertices. We are now going to determine the number of dots in each stage, and then we could see the pattern of the number of dots. Finally, we can generate a simplified formula or general statement for to find the number of dots in each stage.

Triangular numbers are sequence of number that has certain pattern and if we plot it into a diagram it will create a geometrical triangular shape. Each triangular number has its own number of dots and formula. Now, we will start with triangular number first.

Aim: In this task I will consider geometric shapes which lead to special numbers. The simplest example of these are square numbers, 1,4,9,16 which can be represented by squares of side 1,2,3 and 4.

The following diagrams show a triangular pattern of evenly spaced dots. The numbers of dots in each diagram are examples of triangular number (1,3,6,...).

[pic] [pic] [pic][pic]

1 3 6 10 15

Complete the triangular numbers sequence with three more terms.

[pic] [pic] [pic]

21 28 36

S6 = 21 dots

S7= 28 dots

S8 = 36 dots

General statement that represents the nth triangular number in terms of n

Sn = 1/2n2 + 1/2n

Consider stellar (star) shapes with p vertices, leading to p-stellar numbers. The firsr four representations for a star with six vertices are shown in the four stages S1-S4 below. The 6-stellar number at each stage is the total number of dots in the diagram.

[pic] [pic] [pic]

S1 S2 S3 S4

Number of dots in

S1= 1 dot

S2= 13 dots

S3= 37 dots

S4= 73 dots

[pic] [pic]

S5 = 121 dots S6 = 181 dots

Find an expression for the 6-stellar number at stage S7

Sn = an3 + bn2 + cn +d

1. 1 = (ax1)+(bx1)+(c+1)+d

2. 13= 8a + 4b+2c+d

3. 37=27a+9b+3c+d

4. 73=64a+16b+4c+d

• Subs 1 and 2 Subs 2 and three Subs 3 and four

12 = 7a+3b+c 21=19+5b+c 36=37a+7b+c

• Subs red and blue Subs blue and green

12=12a+2b 12=18a+2b

• Subs orange with purple

A= 0 B=6 C= -6 D= 1

General formula for 6-stellar numbers at stage Sn in terms of n

Enter all values of A, B, C and D into one of the equations and simplify it, and here is the final result,

Sn = 6n2-6n+1

6 Stellar number at stage S7

Sn = 6n2-6n+1

-Which n as the number of term, since we are going to find the 7th term, we need to substiture n as 7

S7 = 6(7)2 – 6(7) + 1

= 294 – 42 +1

= 253

Repeating steps above for other values of p (1st)

P = 3

[pic] [pic] [pic] [pic]

S1 S2 S3 S4

|Nth term |Number of dots | |1st term |1 dot | |2nd term...
tracking img