Author 
Topic 

Nadeshiko
New Member
USA
2 Posts 
Posted  09/21/2013 : 00:21:11

Hello! Please help me on this problem. I tried and I still don't know how to do it. I'm sorry but I forgot the work that I tried. I did it in my head. Sorry. Please help me with the problem below.
1 2 3 4 5 6 7 8 9 10 11...
Questions: 1. How many #'s in row 30? 2. What row is #200 in? Where is its position in that row? 3. What # does row 12 start with? 4. Where is #2015? Give row and position. 5. What does row 16 end with?
Patterns: Write at least 3 patterns you notice from this assignment.



Ultraglide
Advanced Member
Canada
299 Posts 
Posted  09/23/2013 : 11:37:38

You need to show your work. 


royhaas
Moderator
USA
3059 Posts 
Posted  09/23/2013 : 12:03:29

I wish teachers would quit assigning these things. All they show is that a teacher has a preconceived notion of a pattern, and shows ignorance of number theory and finite differences. 


TchrWill
Advanced Member
USA
80 Posts 
Posted  09/25/2013 : 16:25:19

quote: Originally posted by Nadeshiko
Hello! Please help me on this problem. I tried and I still don't know how to do it. I'm sorry but I forgot the work that I tried. I did it in my head. Sorry. Please help me with the problem below.
1 2 3 4 5 6 7 8 9 10 11...
Questions: 1. How many #'s in row 30? 2. What row is #200 in? Where is its position in that row? 3. What # does row 12 start with? 4. Where is #2015? Give row and position. 5. What does row 16 end with?
Patterns: Write at least 3 patterns you notice from this assignment.
Are you sure of the pattern you offer? Might it really be
1 2.3 4.5.6 7.8.9.10.
What series do the last digits in each row form? 
Edited by  TchrWill on 09/25/2013 16:27:38 


the_hill1962
Advanced Member
USA
1469 Posts 
Posted  09/25/2013 : 18:02:42

I wondered the same thing, TchrWill. Since Nadeshiko has not replied, I think it might be a troll. The problem is impossible as stated. There is no pattern. Maybe if it is a real person, obviously a lazy one since no work or thoughts on how to start it. So, the three dots could be Nadeshiko's way of saying there are more numbers but I am too lazy to type them.
Maybe the pattern is
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 and so on...
Now, there is something to work with. However, since Nadeshiko has not even commented, I don't think we should discuss the problem further except for showing what the next lines are. 32 33 34 35... ...61 62 63 64 65 66... ...126 127 . . .

Edited by  the_hill1962 on 09/25/2013 18:07:58 


TchrWill
Advanced Member
USA
80 Posts 
Posted  09/27/2013 : 09:51:46

In addition to the obvious pattern identified by the_hill1962, there are
1 2.3 4.5.6 7.8.9.10 11.12.13.14.15 16.17.18.19.20.21...yielding the series of
1.3.6.10.15.21... and 1.2.4.7.11.16...
A stretch might be 1 2. 3 4. 5. 6. 7 8. 9.10.11.12.13.14 15.16.17.18.19.20.21.22.23.24.25 26.27.28.29.30.31.32.33.34.35.36.37.38.39.40.41 42.43.44.45...
The first step is to derive the equation that defines each series.
For example, consider the sequence N = 3, 9, 19, 35, 59, 93...
An expression can be derived enabling the definition the nth term of any finite difference series, one where the nth differences are constant. The expression is a function of the number of successive differences required to reach the constant difference. If the first differences are constant, the expression is of the first order, i.e., N = an + b. If the second differences are constant, the expression is of the second order, i.e., N = an^2 + bn + c. Similarly, constant third differences derive from N = an^3 + bn^2 + cn + d.
Take the following example: n.................1......2.....3.....4.....5......6.......nN.........NN.................3......9.....19....35....59....93.......Nn 1st Diff.............6.....10.....16....24....34 2nd Diff................4.......6......8.....10 3rd Diff....................2......2......2
Using the data points (n1, N1), (n2,N2), (n3,N3), etc., we substitute them into N = an^3 + bn^2 + cn + d as follows: (n1,N1) = (1,3) produces a(1^3) + b(1^2) + c(1) + d = 3 or a + b + c + d = 3 (n2,N2) = (2,9) produces a(2^3) + b(2^2) + c(2) + d = 9 or 8a + 4b + 2c + d = 9 (n3,N3) = (3,19) produces a(3^3) + b(3^2) + c(3) + d = 19 or 27a + 9b + 3c + d = 19 (n4,N4) = (4,35) produces a(4^3) + b(4^2) + c(4) + d = 35 or 64a + 16b + 4c + d = 35 Subtracting each successive pair yields 7a + 3b + c = 6 19a + 5b + c = 10 37a + 7b + c = 16 Again, subtracting each successive pair yields 12a + 2b = 4 18a + 2b = 6
Subtracting these yields 6a = 2 making a = 1/3, b = 0, c = 11/3, and d = 1 resulting in our final expression for the nth term of this series Nn = (n^3)/3 + (11n)/3  1 = (n^3 + 11n  3)/3. Checking it out for the 6th term we have [(6^3) + (66)  3]/3 = [216 + 66  3]/3 = 279/3 = 93.
Now you can explore other derived series. Questions: 1. How many #'s in row 30? 2. What row is #200 in? Where is its position in that row? 3. What # does row 12 start with? 4. Where is #2015? Give row and position. 5. What does row 16 end with?

Edited by  TchrWill on 09/27/2013 13:49:23 


TchrWill
Advanced Member
USA
80 Posts 
Posted  09/29/2013 : 11:05:42

the example shown should look like the following:
n...............1......2.....3.....4.....5......6.......nN N...............3......9.....19....35....59....93.......Nn 1st Diff...........6.....10.....16....24....34 2nd Diff.............4.......6......8.....10 3rd Diff.................2......2......2

Edited by  TchrWill on 09/29/2013 16:17:21 



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