I made this table to show that numbers of dots in each term is equal to the number of dots of that term plus the number of the previous term:
T1:1+2=3
T2:1+2+3 =6
T3:1+2+3+4=10
T4 :1+2+3+4+5=15.
I think it is clear example of arithmetic progression, and to find a general statement I can use the formula from the way we find a sum of certain numbers of an arithmetic sequence:
Tn=n(a1+an)/2 where an =a1 +(n-1)d, so I have
Tn= n(2a+d(n-1))/2 where a represents the number of dots in the first term, and d is the common difference between terms.
I know that d represents the difference between terms, so for triangular sequence it will have value egual one. The number of dots in the first term is 1, so a is egual 1 too.
With knowing all information I can substitute it into formula :
Tn= n(2*1+1(n-1))/2 = n(2+n-1)/2=n(n+1)/2.
Now I have:
Tn= n(n+1)/2.
To make sure that I have the right formula I will substitute some numbers:
T3= 3(3+1)/2=6 T6= 6(6+1)/2=21
T4=4(4+1)/2=10 T7=7(7+1)/2=28
T5=5(5+1)/2=15 T8= 8(8+1)/2=36.
From the checking I have this formula working for every term.
Stellar numbers
Stellar shapes – are the shapes with p vertices leading to p stellar numbers. The first four representations for the star with six verticers are shown in the four stages below( S1- S 4).The 6 stellar number at each stage is total number of dots in the diagram:
S 1 S 2 S 3 S4
To find out the number of dots in each stage up to S6 I’m going to organize the data to figure out the difference between the diagrams:
Unfortunately I can’t see the clear pattern from my first table , so I will do the second one with another details to figure out the difference using the numbers in each layer :
Now I see that there is 12 new dots in each next layer. I have noticed the regularity that there one more layer around each term, and each term has 12 times the previous number of dots. I can represent this sentence by formula: 12*Sn-1.
Again I come back to arythmetic sequence formulas: to find out any term of progression we use an= a1+d(n-1). In this particular case d is difference which eguals 12, and a1 represents by Sn-1.
Substituting my information to formula I have
Sn= . Sn-1. + 12( n-1).
But I ‘ve figured out that this formula doesn’t work for S1 because it has only 1 dote so I know that n is supposed to be > 1.
Now I can check my formula :
S2 = 1+12(2-1)=13
S3= 13+12(3-1)=37
S4=37+12(4-1)=73
It does work so I can find out the number of dots at S5 and S6:
S5 =73+ 12(5-1)= 121
S5
Now I will find the number of dots at term S6 :
S6 = 121+12(6-1)=121+60=181
I organize the data of all stellar shapes below to show the regularity between them:
Therefore, I see that formula will work for S7 but It won’t work for another terms, because you always have to know the previous one.
I assume that I can use the same formula that I used with the triangular numbers but I need always add 1, because of the first term.
Using previous information I have Sn = n(2a+d(n-1))/2 +1
I will use 12 as the first term(a) which is the same as S2 ( if you add 1), so that’s why n is going to be one less than 7 which is 6.
So I can substitute it using calculator : S7 = 6(2*12 +(7-1)12))/2 +1 = 253.
So now I have S7 as 1+12+24+36+48+60+72 = 253, which proves the pattern and appropriate for it.
Because the value of a is S2 instead of the first term , all formula will be equal to S n+1:
Using all the knowledge I have
Sn+1 = n(2*12+12(n-1))/2 +1
Using algebra I simplified it into a formula :
S n+1 = 6n(n+1)+1
Therefore I think we need to represent n as n-1, because this equation is equal to S n+1 , so now I substitute and get this:
Sn+1 =6(n-1)( n-1+1)+1=6n2-6n+1.
This general statement can be applied for 6 stellar numbers
by
kamilaakshulakova (student)
