You didn't notice when you wrote
3-2=1, then 5-3=2, 8-5=3, etc,
That the pattern does not work for the next two terms:
3-2=1, then 5-3=2, 8-5=3, 14-8=6, 19-14=5.
Edwin
In a message dated 1/2/2014 10:50:51 A.M. Eastern Standard Time, devnull@algebra.com writes:
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Oh sorry, I messed up the calculations, of course it should be:
2 3 5 8 12 17 23 30 38 47 etc
so +1,+2,+3,+4,+5,+6,+7,+8,+9,etc
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Hi,
The general term is
Here's what you do. You list
the numbers in a column like this:
2
3
5
12
17
23
30
38
47
Then you make a column out beside that one, putting beside each number the
difference between it and the number just below it, like this:
2 1
3 2
5 3
8 4
12 5
17 6
23 7
30 8
38 9
47
Then out beside that, you make a column of differences of differences:
2 1 1
3 2 1
5 3 1
8 4 1
12 5 1
17 6 1
23 7 1
30 8 1
38 9
47
So it only takes 2 columns to get the differences to all be the same. So that
means you have to assume a 2nd degree polynomial general term of the form:
y = ax²+bx+c
The unknowns will be a,b, and c. So you'll need to substitute three points
(x,y) = (n,an), for n=1,2,3
(1,2), (2,3), (3,5)
2 = a(1)²+b(1)+c
3 = a(2)²+b(2)+c
5 = a(3)²+b(3)+c
That becomes:
a + b + c = 2
4a + 2b + c = 3
9a + 3b + c = 5
Subtract the 1st eq. from the 2nd
to eliminate c
3a + b = 1
Subtract the 2nd eq. from the 3rd
to eliminate c.
5a + b = 2
Then subtract those equations to
eliminate b
2a = 1
a = 1/2
Then substitute in 3a + b = 1
3(1/2) + b = 1
3/2 + b = 1
b = -1/2
Then substitute in
a + b + c = 2
(1/2) + (-1/2) + c = 2
c = 2
So y = ax²+bx+c becomes
y = (1/2)x² - (1/2)x + 2
and you can write that as
y = (x²-x+4)/2
and change the x to n and the y to an
an = (n²-n+4)/2
You can do that with ANY sequence of numbers, but very often it takes more than
just 2 columns to get them all the same. For instance there is a 5th degree
equation that would give you the general term for the sequence you made a
mistake on.
2 1 1 0 2 -8
3 2 1 2 -6
5 3 3 -4
8 6 -1
14 5
19
That takes 5 columns to get to a column
that has "all the same" number. Yes that last column only has 1 number in it,
-8, but "all 'ONE' of it is 'the same'". Hahaha :)
So there is a 5th degree polynomial
y=ax^5+bx^4+cx³+dx²+ex+f
There are 6 unknowns, so you'd have to substitute the six points
(1,2), (2,3), (3,5), (4,8), (5,14), (6,19)
in that and you'd have 6 equations in 6 unknowns. Then you can eliminate f by
subtracting, and get 5 equations in 5 unknowns and then get 4 equations in 4
unknowns, then 3 in 3, and 2 in 2 and finally 1 equation in 1 unknown. Then
substitute back and you will get a very complicated 5th degree general term for
the sequence you posted.
That will always work with any sequence. But the general term will be
complicated if there is no easy pattern.
Happy new year,
Edwin