I can't seem to figure out what the next two terms of this sequence are as I can't figure out the term-to-term rule for it. Can anyone help?
11, 9, 10, 17, 36, 79,....
I don't know if this helps you at all, but I believe it is possible to find a polynomial of at most degree 5 that will fit that sequence, by solving simultaneous linear equations.
I will illustrate the method on a smaller scale. Say the sequence is
11, 9, 10, ...
Then we can find a polynomial of degree 2 that will generate this sequence, of the form f(x) = ax^2+bx+c, as follows:
We know f(1) = 11. Thus, a*(1^2)+b*1+c = 11. Simplifying, a+b+c=11.
We know f(2) = 9. Thus, a*(2^2)+b*2+c = 9. Simplifying, 4a+2b+c=9.
We know f(3) = 10. Thus, a*(3^2)+b*3+c = 10. Simplifying, 9a+3b+c=10.
This is a system of linear equations with three equations and three unknowns. The solution is described by (a,b,c) = (3/2, -13/2, 16). That is, f(x) = (3/2)x^2 - (13/2)x + 16. This generates the sequence
11, 9, 10, 14, 21, 31, ...
Obviously this is not the desired sequence, so a quadratic curve-fitting scheme is not feasible. I would proceed by next trying a cubic polynomial to model the first four terms, and if that doesn't work, a degree 4 polynomial to include the first five terms, and in the worst case, a degree 5 polynomial.
Of course there may be a much simpler explanation of your sequence that I'm blind to...
Hello, nsingh201!
Find the next two terms of this sequence: .
Given a sequence of increasing terms, I take the differences of consecutive terms,
. . then the differences of the differences, and so on . . . hoping to find a pattern.
Here's what I found . . .
. .
Since the 3rd differences seem to be identical to the 2nd differences,
. . the function appears to be exponential.
I also suspect that the 2nd differences are doubling.
So I believe the chart can be continued like this;
. .
And I think the generating function is: .
11, 9, 10, 17, 36, 79, - Wolfram|Alpha
2 Soroban
Nice solution, but how did u get generating function?
Hello, ICanFly!
I was afraid someone would ask . . .LOL!
. . .but how did u get generating function?
We have: .
And we know there is a "doubling" feature in there: .
Since the doubling didn't begin until the second differences.
. . I assumed there was a linear or quadratic involved, too.
So, my first general function was: .
I used the first three terms of the sequence:
. .
Substitute into [4]: .
Substitute into [1]: .
Hence: .
I tested my function up to the 8th term . . . It works!
. . Therefore: .