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New Discovery! 2023 Hottest in over 120 Million Years 2024 and beyond in prophecy Questions This Speech Just Broke the Internet Status: Not Logged In; Sign In United States News See other United States News Articles Title: Do you think this is a reasonable math prbolem for a 4th grader? Source: [None] URL Source: [None] Published: May 24, 2016 Author: me Post Date: 2016-05-24 16:48:35 by no gnu taxes Keywords: None Views: 1790 Comments: 30 John saves $1 on day 1, $3 on day 2, $6 dollars on day 3, 10 dollars on day 4 and so on. Given this pattern, how long will it take John to save $165? This seems like at the very least a 7th grade problem. Post Comment   Private Reply   Ignore Thread   Top • Page Up • Full Thread • Page Down • Bottom/Latest Begin Trace Mode for Comment # 20. #14. To: no gnu taxes (#0) Do you think this is a reasonable math prbolem for a 4th grader? No, it is basically an algebra problem though it can be solved by brute force addition if the student recognizes the pattern and can add. SOSO  posted on  2016-05-24   21:59:02 ET  Reply   Untrace   Trace   Private Reply   #15. To: SOSO (#14) Add what? Fred Mertz  posted on  2016-05-24   22:12:44 ET  Reply   Untrace   Trace   Private Reply   #16. To: Fred Mertz (#15) The sum of numbers starting with 1, 1+2=3, 3+3=6, 6+4=10, 10+5=15, etc. etc. That is an inelegant way for solve the problem that was proposed using just arithmetic. There is a more elegant algebraic approach using mathematics. Both are well above 4th grade level expectations. SOSO  posted on  2016-05-24   23:26:30 ET  Reply   Untrace   Trace   Private Reply   #17. To: Fred Mertz (#16) The sum of numbers starting with 1, 1+2=3, 3+3=6, 6+4=10, 10+5=15, etc. etc. That is an inelegant way for solve the problem that was proposed using just arithmetic. There is a more elegant algebraic approach using mathematics. Both are well above 4th grade level expectations. Before you ask the answer is n(n+1)/2. SOSO  posted on  2016-05-24   23:30:25 ET  Reply   Untrace   Trace   Private Reply   #18. To: SOSO (#17) Before you ask the answer is n(n+1)/2. That gives a digit, for n=9 it would yield 45 (9*10/2), correct for the ninth digit in the sequence, but not the sum of the digits. The formula for the sum is: f(n) = n(n+1)(n+2)/6 For n=9: f(9) = 9(10)(11)/6 f(9) = 990/6 = 165 nolu chan  posted on  2016-05-25   1:29:30 ET  Reply   Untrace   Trace   Private Reply   #20. To: nolu chan (#18) The formula for the sum is: f(n) = n(n+1)(n+2)/6 Why divided by 6? That is not intuitive. Vicomte13  posted on  2016-05-25   9:52:40 ET  Reply   Untrace   Trace   Private Reply   Replies to Comment # 20. #21. To: Vicomte13 (#20) I'm certainly not going to through the algebraic manipulations it would take to form an equation. My guess is you would have to solve it like forming the equation for figuring each period payment to pay off an interest loan. You form a general equation, multiply by a factor, and then subtract the original equation. You eliminate all the middle terms, allowing you come up with an equation based on the number of terms. That is at the very least high school level math. no gnu taxes  posted on  2016-05-25 10:18:05 ET  Reply   Untrace   Trace   Private Reply   #23. To: Vicomte13 (#20) Why divided by 6? That is not intuitive. 1*2*3... Not easy to explain. The sum of the first positive integers 1+2+3 ... (n-2)+(n-1)+(n) for n=100 1+100 2+99 3+98 [...] 98+3 99+2 100+1 50x101 = 5050 50 × 101 = (100/2) × 101 = (100/2) × (100 + 1) Substituting n for 100, we get (n/2) × (n + 1) for the SUM of the first n integers. Or n(n+1)/2 Now let us look at the 4th grade problem (not): (1) 1 1 1 (2) 3 4 12 (3) 6 10 123 (4) 10 20 1234 (5) 15 35 12345 (6) 21 56 123456 (7) 28 84 1234567 (8) 36 120 12345678 (9) 45 165 123456789 For n=9, there is 9*1, 8*2, 7*3, 6*4, 5*5, 4*5, 3*6, 2*7, and 1*9. A fourth grader might just add them up. Each of the numbers 1 thru 9 of the sequence is made by the sum of the first n integers, or n(n+1)/2. For n=4, the sequence number is 10 (1+2+3+4). For n=4, the sum of n=1 thru n=4 is (1) + (1+2) + (1+2+3) + (1+2+3+4) or 1+3+6+10 = 20 1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 + 45 (1) 9*0 + 1 = 1 1 (2) 9*0 + 3 = 3 4 (3) 9*0 + 6 = 6 10 (4) 9*1 + 1 = 10 20 (5) 9*1 + 6 = 15 35 (6) 9*2 + 3 = 21 56 (7) 9*3 + 1 = 28 84 (8) 9*4 + 0 = 36 120 (9) 9*5 + 0 = 45 165 9*16=144 plus 1+3+6 + 1+3+6 + 1 = 21. Total 165. To get the sequence number, multiply n*n+1 (two factors) and divide by 2 (1*2). For n=4, multiply 4 x 5 = 20, divide by 2, and the number is 10. For the sum of all sequence numbers to n, multiply n*n+1*n+2 (three factors) and divide by 6 (1*2*3). For sum n=4, multiply 4 x 5 x 6 = 120, divide by 6, and the sum is 20. g(n) = n(n+1)/2 f(n) is the sum of the numbers 1 through n. So: f(n) = g(1) + g(2) + ... + g(n) And: f(n) = n(n+1)(n+2)/6 nolu chan  posted on  2016-05-25 20:30:39 ET  Reply   Untrace   Trace   Private Reply   End Trace Mode for Comment # 20. Top • Page Up • Full Thread • Page Down • Bottom/Latest

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