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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.12 34 5 6 78 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

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 pre-conceived notion of a pattern, and shows ignorance of number theory and finite differences.

TchrWill

USA
80 Posts

 Posted - 09/25/2013 :  16:25:19 quote:Originally posted by NadeshikoHello! 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.12 34 5 6 78 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 be12.34.5.67.8.9.10.What series do the last digits in each row form? Edited by - TchrWill on 09/25/2013 16:27:38

the_hill1962

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 is12 34 5 6 78 9 10 11 12 13 14 1516 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31and 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 6364 65 66... ...126 127... Edited by - the_hill1962 on 09/25/2013 18:07:58

TchrWill

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 7.8.9.1011.12.13.14.1516.17.18.19.20.21...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. 9.10.11.12.13.1415.16.17.18.19.20.21.22.23.24.2526.27.28.29.30.31.32.33.34.35.36.37.38.39.40.4142.43.44.45...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.......Nn1st Diff.............6.....10.....16....24....342nd Diff................4.......6......8.....103rd 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 yields7a + 3b + c = 619a + 5b + c = 1037a + 7b + c = 16 Again, subtracting each successive pair yields12a + 2b = 418a + 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

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.......nNN...............3......9.....19....35....59....93.......Nn1st Diff...........6.....10.....16....24....342nd Diff.............4.......6......8.....103rd Diff.................2......2......2 Edited by - TchrWill on 09/29/2013 16:17:21
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