The formula for the sum of the first n terms of an arithmetic sequence is

Using this formula, we can obtain a general expression of the 1st triangular sequence.

=

Therefore, the stage n or the nth triangular number is represented by the above equation.

For example, if we substitute n as 2, i.e the 2nd stage, it will be equal to

=

= 3

This holds true as we’ve already witnessed the 2nd stage having three dots.

Now let’s take 9th stage into consideration. It will be =

= 45

This triangular number can be seen below and as you can see, this too, holds true.

S9

Hence we can safely acknowledge that is the general expression for this sequence.

This triangular number pattern has also been observed in the Pascal’s Triangle.

Similar to the triangular numbers, there are stellar numbers or star shaped numbers.

Looking at the above star shaped pattern one can see that there are 1, 13, 37, 73 dots in each successive stage and these number of dots in each stage give us a sequence which we can evaluate further. But again, similar to the triangular sequence, this stellar number sequence is neither arithmetic nor a geometric sequence as there is no common difference or a common ration. Hence we cannot apply the formulas for an arithmetic sequence or a geometric sequence.

But if this pattern is observed closely, one can observe that each term in this sequence is the sum of the first 12(n-1) positive integers.

To demonstrate, let a1 , a2… an signify the number of dots in each term in this sequence.

a1 = 1

a2 = 13 = 1+12

a3 = 37 = 1+12+24

a4 = 73 = 1+12+24+36

Just by observing how this sequence is progressing one can find the next three terms in this sequence by applying the above logic.

a5 = 1+12+24+36+(12 x 4) = 121

a6 = 1+12+(12 x 2)+(12 x 3)+(12 x 4)+(12x5) = 181

a7 = 1+12+(12 x 2)+(12 x 3)+(12 x 4)+(12x5)+(12 x 6) = 253

Thus proving the previous statement, that these terms are the sum of the first 12(n-1) integers.

As proved above, this sub-sequence consists of 12, 24, 26, 48… 12(n-1) terms and as observed, this forms an arithmetic sequence with the common difference as 12 and the first term as 12.

Hence, we will again apply the formula for the sum of the first n terms of the arithmetic sequence to help us obtain the general expression for the stellar sequence.

The formula:-

Thus when substituted

= 6n(n-1)

Hence the nth term or the general expression for the stellar sequence is 1+6n(n-1).

Now, if we substitute n as 3, i.e the stellar number at stage 3 (S3), we get

1+6(3)(3-1)

= 1+18(2)

= 1+36

=37

And we have already established that S3 contains 37 dots.

Now for another example, let’s substitute n as 8, i.e the stellar number at stage 8 (S8).

1+6(8)(8-1)

= 1+48(7)

= 1+336

= 337

And you can see that the expression for S8 holds true as it fits into a stellar shape.

And both these above examples go to show that the general expression is legit.

Let’s consider another stellar sequence, except this time with 5 vertices, i.e with p=5.

Now by observing the above stars, one can see that there are 1, 11, 31, 61 dots in each successive stage and thus giving us another sequence to be evaluated further. Again, like the other stellar sequence, there is neither common difference nor a common ration, thus not making it arithmetic or a geometric sequence. This again does not allow us to apply the formulas specific to an arithmetic sequence or a geometric sequence.

But if you observe this pattern carefully, it is seen that the number of dots in each stage is the sum of the first 10(n-1) positive integers.

To demonstrate, lets substitute b1, b2…bn as the terms in this sequence.

b1 = 1

b2 = 11 = 1+10

b3 = 31 = 1+10+20

b4 = 61 = 1+10+20+30

Thus, the 2nd sequence, which is 10, 20, 30… (n-1) is an arithmetic sequence with the first term as 10 and the common difference as 10.

So to obtain the general expression of the stellar sequence we will again apply the formula for the sum of first n terms of the arithmetic sequence.

The formula:-

Therefore when substituted

= 5n(n-1)

Therefore the nth term or the general expression for this stellar sequence is 1+5n(n-1).

To test the validity of this expression, lets substitute n as 3 in the above equation; stage 3 (S3).

1+5(3)(3-1)

= 1+15(2)

= 1+30

= 31

And it has already been previously established that at stage 3 (S3), the star, contains 31 dots.

For another example, let’s substitute n as 5.

1+5(5)(5-1)

= 1+25(4)

= 1+100

= 101

As can be seen in the star shape below, there are 101 dots in the 5th stage of the stellar sequence.

Consider another stellar sequence, except this time with 7 vertices, i.e with p=7.

Now by observing the above stars, one can see that there are 1, 15, 43, 85 dots in each successive stage and thus giving us another sequence to be evaluated further. Again, like the other stellar sequence, there is neither common difference nor a common ration, thus not making it arithmetic or a geometric sequence. This again does not allow us to apply the formulas specific to an arithmetic sequence or a geometric sequence.

But if you observe this pattern carefully, it is seen that the number of dots in each stage is the sum of the first 14(n-1) positive integers.

To demonstrate, lets substitute c1, c2…cn as the terms in this sequence.

c1 = 1

c2 = 15 = 1+14

c3 = 31 = 1+14+28

c4 = 61 = 1+14+28+42

Thus, the 2nd sequence, which is 14, 28, 42… (n-1) is an arithmetic sequence with the first term as 14 and the common difference as 14.

So to obtain the general expression of the stellar sequence we will again apply the formula for the sum of first n terms of the arithmetic sequence.

The formula:-

Therefore when substituted

= 7n(n-1)

Therefore the nth term or the general expression for this stellar sequence is 1+7n(n-1).

To test the validity of this expression, let’s substitute n as 3 in the above equation; stage 3 (S3).

1+7(3)(3-1)

= 1+21(2)

= 1+42

= 43

And it has already been previously established that at stage 3 (S3), the star, contains 31 dots.

For another example, let’s substitute n as 5.

1+7(5)(5-1)

= 1+35(4)

= 1+140

= 141

As can be seen in the star shape below, there are 141 dots in the 5th stage of the stellar sequence.

We have witnessed above how the general expressions for 5, 6 and 7 values of p have been 1+5n(n-1), 1+6n(n-1), 1+7n(n-1), respectively. It is observed that each p value is used in the general equation.

Hence, we can safely assume the general expression for p stellar numbers for any value of p at stage n (Sn) is

1+pn(n-1).

To demonstrate, let’s take the value of p as 8 and n as 3. That shall give us

1+8×3×(3-1)

=1 + 24(2)

= 1+48

= 49

The above means that the stellar number with 8 vertices at stage 3 will have 49 dots. As can be seen in the stellar shape below, the star has 49 dots.

Let’s take another value of p and n. Let p equal to 10 and n equal to 4.

1+104(4-1)

=1+40(3)

=1+121

= 122.

The above number signifies that the stellar number with 10 vertices at stage 4 will have 122 dots. This can be seen in the star shape below.

That is a very good attempt, but I think there can be many changes made.

The changes that I have made are with the notation of terms in a sequence:

- Small letters
- Every sequence is called by a different alphabet.
- The formula is also modified.

You can do the following:

- Copy the star shapes from the inkscape directly to the word file.
- The dots can be put after you take the printout, with black pen.
- More values can be tested for the validity of the general statement and the same can be proved by the star (diagram)
- Discuss the limitations and scope of the task, variables (p and n) and also the technology that you are using. Any difficulties faced by you. (Put in a positive way).
- Conclude the task very nicely by giving the overall process that you had carried out in this research.
- Also provide footnotes or bibliography of things taken from any resource.
- Note that your work should have a nice flow and should not end suddenly.
- Tabulating your data obtained before making any general statement is always good.
- Overall the work is fine.