How is -1 the next number in this sequence -145, -91, -49, -19?

Kessen

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How is -1 the next number in this sequence -145, -91, -49, -19?
What is the full formula of the quadratic sequence and how is it derived?
So far I have deciphered the first part of the quadratic formula as -6n2 by using the details shown below.
View attachment 15929
Thank you
-
 
How is -1 the next number in this sequence -145, -91, -49, -19?
What is the full formula of the quadratic sequence and how is it derived?
So far I have deciphered the first part of the quadratic formula as -6n2 by using the details shown below.
View attachment 15929
Thank you
Nothing shown above!!
 
How is -1 the next number in this sequence -145, -91, -49, -19?
What is the full formula of the quadratic sequence and how is it derived?
So far I have deciphered the first part of the quadratic formula as -6n2 by using the details shown below.
View attachment 15929
Thank you
-

The first differences are:

-54, -42, -30

And the second difference is constant at -12. So, the next first difference will be -18, which makes the next term -1.

Since the second difference is constant, we will indeed have a quadratic for the \(n\)th term. So, we may use the given values to obtain a system of equations:

[MATH]a_1=a+b+c=-145[/MATH]
[MATH]a_2=4a+2b+c=-91[/MATH]
[MATH]a_3=9a+3b+c=-49[/MATH]
Can you proceed?
 
I cannot proceed. The first part of the quadratic is -6n2 but the latter terms remain unknkown.

For the 5th term in the original sequence, (term a5) where n=5 and we know that a5 = -1
a5= −6n2+ ?n + ? = −1

notes.png
The difference between the expected and acheived terms is as follows

My sequence as detailed above in red (-6n2)-6-24-54-150
Original sequence-145-91-49-19
The difference between my sequence and the original sequence is inconsistant as shown here >-151-115-103-169

 
Last edited:
Using a CAS, I get:

[MATH](a,b,c)=(-6,72,-211)[/MATH]
And so:

[MATH]a_n=-6n^2+72n-211[/MATH]
 
I cannot proceed. The first part of the quadratic is -6n2 but the latter terms remain unknkown.

For the 5th term in the original sequence, (term a5) where n=5 and we know that a5 = -1
a5= −6n2+ ?n + ? = −1

View attachment 15935
The difference between the expected and acheived terms is as follows

My sequence as detailed above in red (-6n2)-6-24-54-150
Original sequence-145-91-49-19
The difference between my sequence and the original sequence is inconsistant as shown here >-151-115-103-169

Your fourth term is wrong; -6(4)^2 = -96, not -150. Also, you added rather than take the differences.

But your method is not the only way to solve the problem. Did you not understand post #3, which gave you a system of equations to solve for the coefficients?
 
First there are an infinite number of perfectly valid continuations of that sequence. To find one you can use the fact that, given n points, there exist a unique polynomial, of degree n-1 or less, passing through those n points.

Here, taking the "x" values to be 0, 1, 2, and 3, we have four points (0, -145), (1, -91), (2, -49), and (3, -19) so we can find a polynomial of degree 3, \(\displaystyle y= ax^3+ bx^2+ cx+ d\). Taking x= 0, y= -145, \(\displaystyle a(0^3)+ b(0^2)+ c(0)+ d= d= -145\). Taking x= 1, y= -91, \(\displaystyle a(1^3)+ b(1^2)+ c(1)+ d= a+ b+ c+ d= -91\), \(\displaystyle a(2^3)+ b(2^2)+ c(2)+ 2= 8a+ 4b+ 2c+ d= -49\), and \(\displaystyle a(3^3)+ b(3^2)+ c(3)+ d= 27a+ 9b+ 3c+ d= -19\).

Since d= -145, we have a+ b+ c- 145= -91 so a+ b+ c= 54, 8a+ 4b+ 2c- 145= -49 so 8a+ 4b+ 2c= 96, and 27a+ 9b+ 3c- 145= -19 so 27a+ 9b+ 3c= 126.

Now we have three linear equations to solve for a, b, and c.
 
Dear all,
After much thought I now understand the method in post #3, it is the first time that I've considered that method.
Also, thanks to your help I can now solve the quadratic equation for the sequence as shown, in part, through the improved table below

Cutt.PNG

Thanks again, Best regards

Kessen
 
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