Math Help - Help with pattern problem

1. Help with pattern problem

Hi,

please could someone assist with the following:
This is the pattern give:
Pattern number 1 2 3 4
Number of dots 2 6 12 20

The question is: Find s formula for the general term

Their solution goes as follows:
The second difference is constant therefore Tn is quadratic - I understand this
therefore Tn = an2+bn+c this i understand
but 2a=a - I cannot understand the basis of this statement
Can anyone enlighten me?

Thanks!

2. Originally Posted by robyn6680
Hi,

please could someone assist with the following:
This is the pattern give:
Pattern number 1 2 3 4
Number of dots 2 6 12 20

The question is: Find s formula for the general term

Their solution goes as follows:
The second difference is constant therefore Tn is quadratic - I understand this
therefore Tn = an2+bn+c this i understand
but 2a=a - I cannot understand the basis of this statement
Can anyone enlighten me?

Thanks!
So the source said 2a = a? Then it's a typo, because this would mean a = 0, which clearly can't be since then we'd have a linear closed form solution.

3. Number of dots 2 6 12 20: oblong numbers.

Does this help?

4. Originally Posted by robyn6680
Hi,

please could someone assist with the following:
This is the pattern give:
Pattern number 1 2 3 4
Number of dots 2 6 12 20

The question is: Find s formula for the general term
note the number of dots rewritten as a product ...

$1 \cdot 2, \, 2 \cdot 3, \, 3 \cdot 4, \, 4 \cdot 5, \, ... \, , \, n(n+1), \, ...$

5. Originally Posted by robyn6680
Hi,

please could someone assist with the following:
This is the pattern give:
Pattern number 1 2 3 4
Number of dots 2 6 12 20

The question is: Find s formula for the general term

Their solution goes as follows:
The second difference is constant therefore Tn is quadratic - I understand this
therefore Tn = an2+bn+c this i understand
but 2a=a - I cannot understand the basis of this statement
Can anyone enlighten me?

Thanks!
You say you understand that $T_n= an^2+ bn+ c$. Since you have 3 numbers to determine, a, b, and c, you need 3 equations. Pick any 3 numbers from the sequence.

Since $T_1= 2$, $a(1^2)+ b(1)+ c= a+ b+ c= 2$.

since $T_2= 6$, $a(2^2)+ b(2)+ c= 4a+ 2b+ c= 6$

Since $T_3= 12$, $a(3^2)+ b(3)+ c= 9a+ 3b+ c= 12$

Subtract the first equation from the second to get (4a+ 2b+ c)- (a+ b+ c)= 3a+ b= 6- 2= 4 or 3a+ b= 4. c has been eliminated.

Subtract the first equation from the third to get (8a+ 3b+ c)- (a+ b+ c)= 7a+ 2b= 12- 2= 10 or 7a+ 2b= 10. Again c has been eliminated. Now we have the two equations 3a+ b= 4 and 7a+ 2b= 10.

Multiply 3a+ b= 4 by 2 to get 6a+ 2b= 8 and subtract that from 7a+ 2b= 10: (7a+ 2b)- (6a+ 2b)= a= 10- 8= 2.

a= 2, not "2a= a".

Now that you know a= 2, use either 3a+ b= 4 or 7a+ 2b= 10 to determine b. Then use any of the original 3 equations to determine c.

As a check, see if those values for a, b, and c give $T_4= 20$.

6. Hi HallsofIvy,
you have a typo there....(9a+3b+c)-(a+b+c)

We end up with a=1, b=1, c=0 leading to skeeter's formula.