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Sequences |
A SEQUENCE is a list of countable objects. A SERIES is the sum of a number sequence.
These objects may be numbers, maybe not.
SEQUENCE EXAMPLES SERIES EXAMPLES
a) 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ... 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + ... =
b) 1, -1, 1, -1, 1, -1, 1, ... 1 + -1 + 1 + -1 + 1 + -1 + 1 + ... =
c) 2, 4, 6, 8, 10, 12, 14, ... 2 + 4 + 6 + 8 + 10 + 12 + 14 + ... =
d) 1, 3, 5, 7, 9, 11, 13, ... 1 + 3 + 5 + 7 + 9 + 11 + 13 + ... =
e) 1/1, 1/2, 1/3, 1/4, ... 1/1 + 1/2 + 1/3 + 1/4 + ... =
f) 1/1, 1/2, 1/4, 1/8, ... 1/1 + 1/2 + 1/4 + 1/8 + ... =
g) 1, 1, 2, 3, 5, 8, 13, 21, ... 1 + 1 + 2 + 3 + 5 + 8 + 13 + ... =
h) 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, ... 1 + 4 + 9 + 16 + 25 + 36 + 49 + ... =
i) A, B, C, D, E, F, G, ...
j) O, T, T, F, F, S, S, E, N, ...
k) penny, nickel, dime, ...
l) Loo, Looloo, Loolooloo, ...
m) Washington, Adams, Jefferson, Madison, ...
1 2 3 4
RANDOM SEQUENCES ...
1, 4, 1, 4, 2, 1, 3, 2, 5, 6, 2, ...
2, 7, 1, 8, 2, 8, 1, 8, 2, 8, 4, 5, 9, 0, 4, 5, 2, 3, ...
3, 1, 4, 1, 5, 9, 2, 6, 5, ...
4, 8, 15, 16, 23, 42, ...
If we index each item in a sequence, then the first item is item 1;
the second, item 2, etc.The nth item in a sequence is called item n. Item n is sometimes called the nth term. Click here for a Sequences Video Demo. SEQUENCE ASSIGNMENTFrom the sequences below, pick one sequence no one has yet picked. Find the next number in the sequence. Write a polynomial formula that finds the nth item in the sequence. Remember polynomials? Polynomials are formulas that have this form: t = an³ +bn² + cn + d. Click here for everything you never wanted to remember about Polynomials. A polynomial, like all formulas, inputs a number and outputs another. Polynomials have degrees, or powers, above the input variable, like n³, and coefficients, which are numbers in front of the input variable, like 2n. Ouch! That's a lot to remember. But you don't need to remember all that to find your formula in this assignment.Create some symbols if you need them. Tell us what they mean. You can write n³ as n^3 or nxnxn or n*n*n. Post sequence, next number, and formula in a comment. You earn 1 point for posting the sequence and the next number. You earn 6 points for the formula. WARNING! Some of you will post what are called recursive algorithms rather than polynomial formulas. You will receive no points for a recursive formula. You will receive points only for a polynomial whose input is the index of the term and whose output is the term. When you begin a problem, you claim it for 1 week only. If you have not worked on the problem for 1 week, you lose your claim, and someone else may claim your problem. Once someone else begins a problem, you cannot claim it for at least 1 week, even if the other person has not completed it, or has completed it incorrectly. You may switch problems if no one has begun working on it. If you do, you relinquish claim on all previous problems you worked on. Since there are not enough problems to go around, after all problems are taken, complain; and a special problem will be given to you in a comment. First come, first served.
latecomers can email me the polynomial for this sequence: 3, 15, 33, 57, 87, ... SOLUTIONS OF EXAMPLES
|
Comments:
From wHolt - 12/14/06 11:36 AM
From HotrodMinivan - 12/13/06 1:13 PM
#15. 1, 3, 7, 15, 31, 63, 127, 255, 511, 1023 . . 2447
v v v v v v v v v v
2 4 8 16 32 64 128 356 712 1424
The difference between any two numbers doubles so the
equation is (t = n2 - n + 2)
From BassLady - 12/13/06 1:12 PM
Let's try this for a formula:
#11 - Sequence: 3, 14, 31, 54, 83, 118, 159, 206 and so on......
Formula: Next in sequence can be found by
X = unknown next sequential number F = final known number in sequence P = number prior to last known number in sequence
X = (f - p) + 6 + f
From wHolt - 11/17/06 1:54 PM
Butterfly - did you examine the other examples that had similar sequences?
SuperDuke had one like yours.
BassLady - I cannot explain it any better than Hypatia...
From BassLady - 11/16/06 6:13 PM
Butterfly,
I feel your pain. I know what it the sequence is and how I got it, but a formula for it?????????
From wHolt - 11/9/06 12:52 PM
email me the polynomial for this sequence:
3, 15, 33, 57, 87, ...
From wHolt - 11/8/06 10:43 AM
It also may help to read your fellow students' comments.
From Butterfly - 11/7/06 1:34 PM
I am trying #19. I am having much difficulty. The above example given is not working on this one. 1, -1/6, 1/120, -1/5040 1.166666667, -.175, .008531746 1.341666667, -.183531746 1.525198413 I am very lost. I think from looking at the numbers that 6 will go into each of them. I just have no formula. |
From wHolt - 11/7/06 11:33 AM
| # | Alias | Date Started |
| 1 | Kathi | 10/25 |
| 2 | Boki | 10/25 |
| 3 | Taurus | 10/29 |
| 4 | David | 11/05 |
| 5 | Hypatia | 10/25 |
| 6 | Slick | 10/31 |
| 7 | Poovey | 11/02 |
| 8 | 7Iron | 10/31 |
| 9 | JooJoo | 10/29 |
| 10 | Pac | 11/03 |
| 11 | BassLady | 11/04 |
| 12 | Sunshine | 11/06 |
| 13 | Pringle | 10/31 |
| 14 | Houdini | 10/29 |
| 15 | Hotrod | 11/04 |
| 16 | Melewen | 11/02 |
| 17 | SuperDuke | 11/05 |
| 18 | Harkar | 11/06 |
| 19 | Butterfly | 11/09 |
| 20 | Bravo | 11/07 |
From wHolt - 11/7/06 11:32 AM
You must work your claim to own it.
From wHolt - 11/7/06 11:31 AM
Can you express your sequence a little differently so that n starts at 1?
From DirtyBird - 11/6/06 8:12 PM
Mr. Holt,
Can you post a few more problems. I think all the problems have been accounted for. Thanks.
From Bravo - 11/6/06 6:44 PM
I claim number 20. 1, -1/2, 1/24, -1/720, ... the next number is 1/40,320
The formula is (-1)^n/(n+1)!
n= 1, 2, 3....
From Poovey - 11/6/06 6:15 PM
#7
5, 15, 37, 77, 141, 235 is the next number
FORMULA N^3 +3N+1
From David - 11/6/06 2:13 PM
#4 2,4,8,14,22,32.....next number is 44
formula
t=n2-n+2
From wHolt - 11/6/06 2:04 PM
From Sunshine - 11/6/06 8:57 AM
#12
2, 5, 10, 17, 26, 37, 50, 65, 82, 101, ... 122
The polynomial formula is t(n)= (n^2)+1Test:
n=4 :: t(4)=(16)+1=17
n=5 :: t(5)=(5^2)+1=25+1
n=6 :: t(6)= (6^2)+1=36+1
All three check.
From wHolt - 11/6/06 12:06 AM
From David - 11/5/06 11:50 PM
From Tiger - 11/5/06 9:02 PM
From Harkar - 11/5/06 3:09 PM
#18 1, 1/2, 1/6, 1/24, 1/120, 1/720, ...
The next number is 1/5040.
1 / n!(n factorial)
Examples:
n(1) = 1/1 = 1
n(2) = 1/1*2 = 1/2
n(3) = 1/1*2*3 = 1/6
n(4) = 1/1*2*3*4 = 1/24
n(5) = 1/1*2*3*4*5 = 1/120
n(6) = 1/1*2*3*4*5*6 = 1/720
From SuperDuke - 11/4/06 2:23 PM
I'll do #17
1,2,6,24,120,720,5040,40320,362880,3628800, the next is 39916800
The set of numbers for problem #17 are the factorial products of the index. The factorial
symbol is denoted as "!". If the index is "n" then n! would represent the product of all the
integers from 1 to n.
Example: 3! = 1*2*3 = 6
Example: 11! = 1*2*3*4*5*6*7*8*9*10*11 = 39916800
From wHolt - 11/4/06 1:52 PM
Hotrod - your formula is recursive.
That is, we need to know the previous term before we can use yours.
I want one where I can plug in 1000 and know what the 1000th term is,
without having to backtrack through 999 previous terms.
From HotrodMinivan - 11/4/06 10:37 AM
15. 1, 3, 7, 15, 31, 63, 127, 255, 511, 1023, 2447 is the next number
ν ν ν ν ν ν ν ν ν ν
2 4 8 16 32 64 128 356 712 1424 . . .(n2)
The difference between each term doubles each time
Formula: t+(n2)
From BassLady - 11/3/06 7:58 PM
#11 3, 14, 31, 54, 83, 118, ...
The next number is 159. I know why - but................I can't write the formula yet.
I'm still working on the rest of the problem. I reserve for a week.
From Pac - 11/3/06 5:13 PM
More on #10:
1, 4, 11, 22, 37, 56, ...
The next numbers are 79, 106, 137, ...
The polynomial formula is t(n)=2(n^2)-3n+2
Test:
n=2 :: t(2)=2(2^2)-3(2)+2 = 2(4)-6+2 = 8-6+2 = 4
n=6 :: t(6)=2(6^2)-3(6)+2 = 2(36)-18+2 = 72-18+2 = 56
n=9 :: t(9)=2(9^2)-3(9)+2 = 2(81)-27+2 = 162-27+2 = 137
All three check.
From Houdini - 11/3/06 1:37 PM
I'm forfeiting my rights to number 17, instead I claim number 14.
1, 16, 81, 256, 625, 1296, 2401, 4096, 6561, 10000, . . .
The next number in the sequence is 14641.
Formula: n^4 or nXnXnXn or n*n*n*n
Test:
1^4=1
3^4=81
22^4=234256
From Pac - 11/2/06 5:33 PM
Just to hold for a week...
#10: 1, 4, 11, 22, 37, 56, ...
The next number is 79
From wHolt - 11/2/06 11:03 AM
From Melewen - 11/2/06 8:23 AM
Oops, no, my formula is
3^n
From Harkar - 11/1/06 6:40 PM
#5
2,6,22,56,114,...
202 is the next number
From Poovey - 11/1/06 6:28 PM
#7
5, 15, 37, 77, 141, 235 is the next number
From Melewen - 11/1/06 4:45 PM
#16. 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, ...
Next number in sequence: 177147
Formula: 3n
From wHolt - 11/1/06 12:03 PM
If you find your formula, you will not need me to verify it.
All the numbers in the sequence will verify it.
From Slick - 10/31/06 4:22 PM
#6 14, 22, 32, 44, 58, 74 is next number
From Pringle - 10/31/06 2:55 PM
13. 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, ...
next number: 1331
fomula : n^3
T(n)= n^3
T(1)= 1^3 =1
T(2)= 2^3= 8
T(3)= 3^3 =27
From 7Iron - 10/31/06 1:10 PM
#8 1,5,12,22,35
1,5,12,22,35,51,70
1,2, 3, 4, 5, 6, 7..n
formula = 1.5(n)^2 - .5(n)
test
t(2) = 1.5(2)^2 - .5(2)
5
t(3) = 1.5(3)^2 - .5(3)
12
t(7) = 1.5(7)^2 - .5(7)
70
From wHolt - 10/31/06 9:49 AM
From Capricorn - 10/29/06 9:53 PM
2, 5, 10, 17, 26, 37, 50, 65, 82, 101, ... 122...n
1 2 3 4 5 6 7 8 9 10 ... 11...n
From Taurus - 10/29/06 9:45 PM
#3. 1, 2, 4, 7, 11, 16, ... n ... 22
1 2 3 4 5 6 ... n ... 7
From Houdini - 10/29/06 2:48 PM
#17: 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, is mine.
From JooJoo - 10/29/06 10:19 AM
#9. 3, 8, 13, 18, 23
The next number would be 28
So it would go: 3, 8, 13, 18, 23, 28,.....,..... because the numbers have difference of 5 in between them
t(n)= 3, 8, 13, 18, 23, n+5,.....
t(n)= 3 + 5(n-1)
t(n)= 5n-2
From wHolt - 10/28/06 2:01 PM
because the next term depends on the previous one.
Keep looking for a polynomial that does not depend on knowing the previous term.
From TBird - 10/27/06 10:09 PM
Check:
t(1)= 2(1)+2 = 4
t(2)= 2(2)+4 = 8
t(3)= 2(3)+8 = 14
t(4)= 2(4)+14 = 22
t(n)= 2(n)+t = 2n+t
From wHolt - 10/27/06 12:12 PM
Everyone should have enough examples now to do one.
From Kathi - 10/27/06 6:07 AM
#2. #1. 3, 9, 15, 21, 27, 33, 39, 45,....
let t(n)=3, 9, 15, 21, 27, 33, 39, 45,……. ...
The difference between each term is 6.
Formula is:
t(n) = 3 + 6(n-1) for n=1, 2, 3, 4,.....
check:
n=1
t(1) = 3 + 6(1-1) = 3 + 6*0 = 3
n=2
t(2) = 3 + 6(2-1) = 3 + 6*1 = 9
n=3
t(3) = 3 + 6(3-1) = 3 + 6*2 = 15
t(n) = 3 + 6(n-1) or simplified: t(n)= 6n-3
From wHolt - 10/26/06 10:29 AM
That does not mean that you must use it;
as long as you find your polynomial you earn the 7 points.
From wHolt - 10/26/06 10:27 AM
If you have () in your formula simplify it the same way. Thanks.
From Boki - 10/25/06 11:02 PM
t(n)=2+5(n-1)
simplifyed:
t(n)=5n-3
From wHolt - 10/25/06 9:06 PM
From Boki - 10/25/06 6:03 PM
#2. 2, 7, 12, 17, 22, 27, 32, 37,…….
let t(n)=2, 7, 12, 17, 22, 27, 32, 37,……. ...
The difference between each term in sequence is 5.
Formula is:
t(n)= 2+ 5(n-1) for n=1, 2, 3, 4, ……….
Check:
n=1
t(1)= 2 + 5(1-1) = 2 + 5*0 = 2
n=2
t(2)= 2 + 5(2-1) = 2 + 5*1 = 2 + 5 = 7
t(3)= 2 + 5(3-1) = 2 + 5*2 = 2 + 10 = 12
:
:
:
n=n
t(n)= 2+ 5(n-1)
From Hypatia - 10/25/06 12:49 PM
I found some help for #5: 2, 6, 22, 56, 114, ...
at http://mcraefamily.com/MathHelp/GeometryPointsFindPolynomialCoefficients.htm
1. I took differences between each term like this:
2 6 22 56 114
4 16 34 58
12 18 24
6 6
till I got a constant difference of 6.
2. It took 3 lines of successive differences to get 6's.
This meant that my formula was a degree 3 polynomial, and so required an n³.
3. Next I divided the constant difference 6 by the number of ways to arrange 3 things.
According to the 3 lady liar puzzle, there are 6 ways to arrange 3 things.
So 6/6 = 1, and 1 is the coefficient of the n³ term.
4. Next I subtracted n³ from each term t:
n t - n³
1 2 - 1³ = 1
2 6 - 2³ = -2
3 22 - 3³ = -5
4 56 - 4³ = -8
5 114 - 5³ = -11
5. Then I found the differences between
1 -2 -5 -8 -11
-3 -3 -3 -3
to be a constant -3.
6. Since this took only 1 step to find the constant difference,
there must be an n¹ first degree term in the polynomial.
And since the number of ways to arrange 1 thing is 1,
I divided -3 by 1: -3/1 = -3.
This meant that -3n was the next term in my sought after polynomial.
7. Next, I repeated step 4, and subtracted n³ and -3n from each term:
n t - n³ - -3 n
1 2 - 1³ - -3(1) = 4
2 6 - 2³ - -3(2) = 4
3 22 - 3³ - -3(3) = 4
4 56 - 4³ - -3(4) = 4
5 114 - 5³ - -3(5) = 4
8. Immediately the constant difference of 4 is revealed.
This means add 4 to my polynomial, and I am finished:
t = n³ -3n + 4
9. Check:
when n=5, t = 5³ -3(5) + 4 = 114
when n=6, t = 6³ -3(6) + 4 = 202
when n=9, t = 9³ -3(9) + 4 = 706
Also, I picked Lucida Console (under Font family) to line up my numbers.
You might like it too.
From wHolt - 10/25/06 12:39 PM
3, 9, 15, 21, 27, 33,...,n + 6, ... [this is the t row]
1 2 3 4 5 6,...,7 + 6=13, ... [this is the n row]
But when n=3, t=15, not 3+6=9
Boki - same goes for your formula.
From Boki - 10/25/06 7:28 AM
#2.
2, 7, 12, 17, 22, 27, 32, 37,……., n+5, ...
From Kathi - 10/25/06 6:34 AM
33 is the next number
#1. 3, 9, 15, 21, 27, 33,...,n + 6, ...
Last Modified 12/18/06 5:33 PM
Look how Hotrod is trying to do it.
HotRod - if n=3, then t should equal 7, but it equals 8. Fix it.