im having trouble with finding the equation to this number sequence. it seems to me there must be some number thats out of place and thats why i cant get it, but the teacher insists that these are the correct numbers: 3, 6, 9, 18, 39, _, _, _. can someone help me find the pattern ASAP
number sequence
- Thread starter Joe05
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78, 141, 234 works with constant 3rd differences (6)
79, 147, 255 works with constant 5th differences (1)
77, 135, 213 works with constant 5th differences (-1)
68, 81, 24 works with constant 5th differences (-10)
I could go on, of course.
What is THE pattern? There isn't such a thing. There are INFINTELY many. If you find one you can explain, and it gets marked wrong, please go ask for your points back. If you can explain it, you should get credit. Guessing what someone else wants is not a worthy criterion for awarding points.
33.99289, 2.022824, 0.000637 works with constant 4th ratios. (0.457031)
Really, INFINITELY MANY!
79, 147, 255 works with constant 5th differences (1)
77, 135, 213 works with constant 5th differences (-1)
68, 81, 24 works with constant 5th differences (-10)
I could go on, of course.
What is THE pattern? There isn't such a thing. There are INFINTELY many. If you find one you can explain, and it gets marked wrong, please go ask for your points back. If you can explain it, you should get credit. Guessing what someone else wants is not a worthy criterion for awarding points.
33.99289, 2.022824, 0.000637 works with constant 4th ratios. (0.457031)
Really, INFINITELY MANY!
- Joined
- Apr 12, 2005
- Messages
- 11,339
Your teacher is being arbitrary/
Your friend is believing it.
You have a desire to get the problem "right", so you believe it, too.
This is also perfectly good: 3, 6, 9, 18, 39, -2, -2, -2 -- It's not as smooth, but certainly can be explained.
You must become acquainted with the "Difference". Simply subtract them all from each other in some sequence.
6-3 = 3
9-6 = 3
18-9 = 9
39-18 = 21
That constitutes all the 1st Differences
Do it again.
3-3 = 0
9-3 = 6
21-9 = 12
That is the known 2nd differences.
You do the rest, but don't you DARE let that teacher get away with the concept that there is ONLY ONE possible result. It's just wrong.
Your friend is believing it.
You have a desire to get the problem "right", so you believe it, too.
This is also perfectly good: 3, 6, 9, 18, 39, -2, -2, -2 -- It's not as smooth, but certainly can be explained.
You must become acquainted with the "Difference". Simply subtract them all from each other in some sequence.
6-3 = 3
9-6 = 3
18-9 = 9
39-18 = 21
That constitutes all the 1st Differences
Do it again.
3-3 = 0
9-3 = 6
21-9 = 12
That is the known 2nd differences.
You do the rest, but don't you DARE let that teacher get away with the concept that there is ONLY ONE possible result. It's just wrong.
Hello, Joe05!
Take the difference of consecutive terms,
then take the differences of the differences, and so on.
\(\displaystyle \begin{array}{cccccccccc}\text{Sequence} & 3 && 6 && 9 && 18 && 39 \\ \text{1st differences} && 3 && 3 && 9 && 21 \\ \text{2nd differences} &&& 0 && 6 && 12 \\ \text{3rd differences} &&&& 6 && 6 \end{array}\)
I suspect that the third differences are constant.
. . Hence, I assume that the generation function is of the third degree ... a cubic.
The general cubic function is: .\(\displaystyle f(n) \;=\;an^3 + bn^2 + cn + d\)
Use the first four terms of the sequence to create a system of equations:
. . \(\displaystyle \begin{array}{ccccccc}f(1) = 3: & a + b + c + d &=& 3 & [1] \\ f(2) = 6: & 8a + 4b + 2c + d &=& 6 & [2] \\ f(3) = 9: & 27a + 9b + 3c + d &=& 9 & [3] \\ f(4) = 18: & 64a + 16b + 4c + d &=& 18 & [4] \end{array}\)
. . \(\displaystyle \begin{array}{ccccccc}\text{Subtract [2] - [1]:} & 7a + 3b + c &=& 3 & [5] \\ \text{Subtract [3] - [2]:} & 19a + 5b + c &=& 3 & [6] \\ \text{Subtract [4] - [3]:} & 37a + 7b + c &=& 9 & [7] \end{array}\)
. . \(\displaystyle \begin{array}{ccccccc}\text{Subtract [6] - [5]:} & 12a + 2b &=& 0 & [8] \\ \text{Subtract [7] - [6]:} & 18a + 2b &=& 6 & [9] \end{array}\)
. . \(\displaystyle \text{Subtract [9] - [8]: }\;6a \:=\:6 \quad\Rightarrow\quad \boxed{a \:=\:1}\)
\(\displaystyle \text{Substitute into [8]: }\;12(1) + 2b \:=\:0 \quad\Rightarrow\quad \boxed{b \:=\;-6}\)
\(\displaystyle \text{Substitute into [5]: }\;7(1) + 3(\text{-}6) + c \:=\:3 \quad\Rightarrow\quad \boxed{c\:=\:14}\)
\(\displaystyle \text{Substitute into [1]: }\;1 - 7 + 14 + d \:=\:3 \quad\Rightarrow\quad \boxed{d \:=\:-6}\)
\(\displaystyle \text{Therefore: }\;\boxed{f(n) \:=\:n^3 - 6n^2 + 14n - 6}\)
Take the difference of consecutive terms,
then take the differences of the differences, and so on.
\(\displaystyle \begin{array}{cccccccccc}\text{Sequence} & 3 && 6 && 9 && 18 && 39 \\ \text{1st differences} && 3 && 3 && 9 && 21 \\ \text{2nd differences} &&& 0 && 6 && 12 \\ \text{3rd differences} &&&& 6 && 6 \end{array}\)
I suspect that the third differences are constant.
. . Hence, I assume that the generation function is of the third degree ... a cubic.
The general cubic function is: .\(\displaystyle f(n) \;=\;an^3 + bn^2 + cn + d\)
Use the first four terms of the sequence to create a system of equations:
. . \(\displaystyle \begin{array}{ccccccc}f(1) = 3: & a + b + c + d &=& 3 & [1] \\ f(2) = 6: & 8a + 4b + 2c + d &=& 6 & [2] \\ f(3) = 9: & 27a + 9b + 3c + d &=& 9 & [3] \\ f(4) = 18: & 64a + 16b + 4c + d &=& 18 & [4] \end{array}\)
. . \(\displaystyle \begin{array}{ccccccc}\text{Subtract [2] - [1]:} & 7a + 3b + c &=& 3 & [5] \\ \text{Subtract [3] - [2]:} & 19a + 5b + c &=& 3 & [6] \\ \text{Subtract [4] - [3]:} & 37a + 7b + c &=& 9 & [7] \end{array}\)
. . \(\displaystyle \begin{array}{ccccccc}\text{Subtract [6] - [5]:} & 12a + 2b &=& 0 & [8] \\ \text{Subtract [7] - [6]:} & 18a + 2b &=& 6 & [9] \end{array}\)
. . \(\displaystyle \text{Subtract [9] - [8]: }\;6a \:=\:6 \quad\Rightarrow\quad \boxed{a \:=\:1}\)
\(\displaystyle \text{Substitute into [8]: }\;12(1) + 2b \:=\:0 \quad\Rightarrow\quad \boxed{b \:=\;-6}\)
\(\displaystyle \text{Substitute into [5]: }\;7(1) + 3(\text{-}6) + c \:=\:3 \quad\Rightarrow\quad \boxed{c\:=\:14}\)
\(\displaystyle \text{Substitute into [1]: }\;1 - 7 + 14 + d \:=\:3 \quad\Rightarrow\quad \boxed{d \:=\:-6}\)
\(\displaystyle \text{Therefore: }\;\boxed{f(n) \:=\:n^3 - 6n^2 + 14n - 6}\)
No need to calculate a,b,c,d; can be done directly this shorter way;soroban said:
take the 1st number of each line {3},{3},{0},{6} and apply like this:
{3} + {3}[(n-1)/1] + {0}[(n-1)(n-2)/(1*2)] + {6}[(n-1)(n-2)(n-3)/(1*2*3) *************
= (3) + (3n - 3) + (0) + (n^3 - 6n^2 + 11n - 6)
= n^3 - 6n^2 + 14n - 6
************ this is a pattern; carries on similarly if more than 4 "lines".