Secondly I can see that the number of squares in the pattern can
be found out by taking the odd numbers from 1 onwards and adding
them up (according to the sequence). We then take the summation
(Σ) of these odd numbers and multiply them by two. After doing this
we add on the next consecutive odd number to the doubled total.
I that I have this information, I can construct a table to find a formula for the numbers. Just before that I will construct a table of the differences found in between the sequence results.
As can be seen, a clear similar number can be seen within the 2nd difference. This will cause the nth formula to have a n2 in it.
I can also form an equation through the trial and improvement method.
1) 2(n -1) (n - 1) + 2n - 1
2) 2(n2 - 2n + 1) + 2n - 1
3) 2n2 - 4n + 2 + 2n - 1
4) 2n2 - 2n + 1
Therefore my final equation is:
2n2 - 2n + 1
Proving My Equation and Using it to Find the Number
of Squares in Higher Sequences
I can deduct from this table and the formulas that the nth term formula can be 2n2 + 2n + 1. There is one problem with this formula. If we take the first sequence to be one square, the formula will not work. Therefore to allow the first term to be 1, I will use the formula 2n2 - 2n + 1. This formula does not change any of the other sequence results.
I will now prove my equation in a number of sequences, including higher sequences that I yet have to explore.
Sequence 3:
1. 2(32) - 6 + 1
2. 2(9) - 6 + 1
3. 18 -5
4. = 13
The formula when applied to sequence 3 appears to be
successful.
Sequence 5:
1. 2(52) - 10 + 1
2. 2(25) - 10 + 1
3. 50 - 10 + 1
4. 50 - 9
5. = 41
Successful
Sequence 6:
1. 2(62) - 12 + 1
2. 2(36) - 12 +1
3. 72 - 12 + 1
4. 72 - 11
5. = 61
Successful
Sequence 8:
1. 2(82) - 16 + 1
2. 2(64) - 16 + 1
3. 128 - 16 + 1
4. 128 - 15 = 113
Successful
The formula I found seems to be successful, as I have shown on the
previous page. I will now use the formula to find the number of squares in a higher sequence.
So now I wil use the formula 2n2 - 2n + 1 to try and find
the number of squares contained in sequence 20.
Sequence 20:
2 (202) - 40 + 1
2(400) - 40 + 1
800 - 40 + 1
800 - 39
= 761
Instead of illustrating the pattern I am going to use the method
I used at the start of this piece of coursework. The method in which
I used to look for any patterns in the sequences. I will use this
to prove the number of squares given by the equation is correct.
As shown below:
2(1+3+5+7+9+11+13+15+17+19+21+23+25+27+29+31+33+35+37) + 39 = 761
I feel this proves the equation fully for a 2d sequence.
3D sequence
The sequence for the 3d sequence is as follows. I built the pattern to get the no of blocks in the first few patterns.
Now I will try to establish the number of squares in a 3d pattern. As stated before the equation for establishing the 2d pattern is:
2n2 -2n+1
This gives the pattern:
For the 3d sequence, I have discovered that the sequence follows a triangular pattern. In this symmetrical pattern, I've found the value of the second sequence to be twice that of the first + the second value.
Thus à 2*1 + 5 à 7
This pattern continues for the third sequence as:
Twice that of the first value+ twice that of the second value + third value
Thus à 2*1 + 2*5 + 13 à 25
Hence, it can be seen that for an nth term, it is the summation of twice [2n2 -2n+1] for the values of n from 1 to n-1.
This can be stated as:
n-1
Σ 2(2n2 -2n+1) + the nth term
1
Hence à {Σ 2(2n2 -2n+1)} + 2n2 -2n+1} Visit coursework fc in fc fo fc for fc more writing fc Do fc not fc redistribute
Now that I have created a formula, I will prove that it works by checking it out on the sequences.
{Σ 2(2n2 -2n+1)} + 2n2 -2n+1}
For my first proof I will use the third sequence pattern. N=3
N = {Σ 2(2n2 -2n+1)} + 2n2 -2n+1
The summation function Σ for the range n=1 to n=n-1, works as follows:
N={2+10} +13
N=25
Correct
I will now use this formula to try for the next sequence
N=4
Sequence 4
N = {Σ 2(2n2 -2n+1)} + 2n2 -2n+1
N={2+10+26}+25
N=63
Successful
Sequence 5
N=5
N={2+10+26+50}+41
N=129
Successful
The formula I found seems to be successful, as I have shown two pages ago. I will now use the formula to find the number of squares in a higher sequence.
Sequence 6
N=6
N={2+10+26+50+82}+61
N=231
Successful
Instead of illustrating the pattern I am going to use the method I used at the start of this piece of this part of the coursework. The method in which I used the triangle to work out the number in the sequence:
(1+5+13+25+41+61+41+25+13+5+1)= 231.
This therefore proves that my sequence is correct , and thus proved.
Conclusion
I have made a number of conclusions from the investigation I have carried out.
Firstly I have found out that the equation used in the 2D pattern was a quadratic. This can be proven through the fact that the 2nd difference was a constant, which is a necessary element of any quadratic. Also as the 2nd difference between the no of blocks was 4, Therefore I knew that the quadratic equation had to include a 2 in it. This allowed me to find out the 2D nth term formula
For the 3D equation, I have discovered that the pattern follows a symmetrical triangle consisting of the numbers from 2D. Thus by using the formula 2n2 -2n+1 from the 2D I have been able to use summation to work out the value of the nth number. The summation function allowed me to sum the results of [2n2 -2n+1] from n=1 to n= n-1. Using this result and combining it with the value of [2n2 -2n+1] at n, I have worked out the number of block squares.
a51m. Thus, we can say that whilst this represents a progression, in the end we have come no closer to any "real" knowledge.
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