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Question 1156464: What is the general formula for the sequence: 0 1 4 10 20 35 56
Found 2 solutions by Boreal, greenestamps:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! the difference between adjacent ones is 1/3/6/10/15/21. The next difference will be seven more from 21 or 28 and the term will ver 56+28, or 84.
The differences are
0
----1
1-------2
----3-------1
4-------3
----6-------1
10------4
----10------1
20-----5
----15-----1
35-----6
----21----1
56
This will be a cubic polynomial, because the third differences are equal: an^3+bn^2+cn+d=0
a=the third difference or 1
12a+2b is first second difference or 2
7a+3b+c=u2-u1 or 2-1 or 1
and a+b+c+d= u1, which is 0
6a=1 so a =(1/6)
12a+2b=2, and therefore 2b=0
7a+3b+c=1 and therefore (7/6)-c=1 and c=-(1/6)
a+b+c+d=0 and d will equal 0 as well.
The general formula is (1/6)n^3-(1/6)n
don't factor it out. For n=1 to 6, the numbers will be the sequence
when n=8 it will be (1/6)512-(8/6). that is (512-8)/6=84.
Sorry I was on autopilot. Hope that helps.
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
The differences between terms form a logical pattern, so we can find a polynomial that produces the given sequence.
Here is a display showing that the sequence produces a constant third difference:
0 1 4 10 20 35 56 <-- original sequence
1 3 6 10 15 21 <-- first differences
2 3 4 5 6 <-- second differences
1 1 1 1 <-- third differences
The constant third difference tells us the sequence can be produced by a polynomial of degree 3: 
If you know some basic differential calculus, that is because the n-th derivative of a polynomial of degree n is a constant.
At this point, we could make a system of four equations in the four variables a, b, c, and d using the first four terms of the sequence:
t(1): 
t(2): 
t(3): 
t(4): 
Finishing the problem by that path is a good exercise. It would be well worth your time and effort to do that and see that you get the right polynomial.
However, this example is a good one to show that often there is a faster and easier way to find the polynomial.
Let's go back to that earlier mention about derivatives of a polynomial.
Given a polynomial of degree 3 with leading coefficient 1, the third derivative is 3! = 6:
function: x^3+....
1st derivative: 3x^2+...
2nd derivative: 6x+...
3rd derivative: 6
Since in this problem the constant 3rd difference is 1, we know that the leading coefficient of the polynomial that produces the sequence is 1/6.
So one thing we could do is compare the terms of the given sequence to the terms of the sequence  .
But that is clearly going to have us working with fractions, which always makes our calculations slower and more difficult.
So what we can do now is multiply every term in the given sequence by 6; that will give us a constant 3rd difference of 1.
And now we can compare the terms of the new sequence to the terms of the sequence  .
term # 6 times given term x^3 difference
-------------------------------------------------
1 0 1 -1
2 6 8 -2
3 24 27 -3
4 60 64 -4
5 120 125 -5
6 210 216 -6
7 336 343 -7
This analysis shows us that the sequence of 6 times the given terms is given by the formula  .
And so the polynomial formula for the given sequence is
CHECK:
t(1): 
t(2): 
t(3): 
t(4): 
t(5): 
t(6): 
t(7): 
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