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May 6th, 2015, 06:01 PM  #1 
Senior Member Joined: Jan 2015 From: USA Posts: 107 Thanks: 2  Let's see if someone can solve this weird sequence
Out of the irreducible fractions smaller than 50 with denominator=6, find the sum. So I discovered that the numerators are: 1, 5, 7, 11, 13, 17, 19... The difference between them are 4,2,4,2,4,2... I don't know how to solve these types of sequences. I also noticed that: first term+second term= 1 third term+fourth term= 3 fifth+sixth= 5 ...and so on, generating an arithmetic sequence: 1,3,5,7... But I'm not sure how to proceed. Also, if you know a method for solving these types of sequences, I'd be delighted to learn it! Thanks heaps. Last edited by skipjack; May 6th, 2015 at 07:39 PM. 
May 6th, 2015, 06:30 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond 
$\displaystyle \sum_{k=1}^{n}(2k1)=n^2$
Last edited by greg1313; May 6th, 2015 at 08:26 PM. 
May 6th, 2015, 07:47 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,942 Thanks: 2210 
I think you intended $\displaystyle \sum_{k=1}^{n}(2k1)=n^2\!$.

May 6th, 2015, 08:30 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,942 Thanks: 2210 
The nth numerator is (3(2n  1) + (1)^n)/2 (which is 3n  2 if n is odd and 3n  1 if n is even), but the key fact is that the first n pairs of fractions total n² (because the kth pair sums to 2k  1). 
May 6th, 2015, 09:44 PM  #5  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Hello, matisolla! Quote:
There are several ways to find this sum. Here is one of them. Find the sum of the numerators. The sum of the first 50 odd numbers is Therefore:  
May 8th, 2015, 03:10 AM  #6 
Senior Member Joined: Jan 2015 From: USA Posts: 107 Thanks: 2  
May 8th, 2015, 05:09 AM  #7 
Global Moderator Joined: Dec 2006 Posts: 20,942 Thanks: 2210 
It's an arithmetic progression (A.P.) with first term 1 and last term n1. By pairing the first term with the last, the second with the second last, etc., the average value of the terms is n, so the value of the sum is n × n = n².


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