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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: 7,929 Thanks: 1124 Math Focus: Elementary mathematics and beyond $\displaystyle \sum_{k=1}^{n}(2k-1)=n^2$ Thanks from matisolla Last edited by greg1313; May 6th, 2015 at 08:26 PM.
 May 6th, 2015, 07:47 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,469 Thanks: 2038 I think you intended $\displaystyle \sum_{k=1}^{n}(2k-1)=n^2\!$. Thanks from greg1313
 May 6th, 2015, 08:30 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,469 Thanks: 2038 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). Thanks from matisolla
May 6th, 2015, 09:44 PM   #5
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Hello, matisolla!

Quote:
 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 . . .

There are several ways to find this sum.
Here is one of them.

Find the sum of the numerators.

$\begin{array}{ccccc}\text{W\!e want:} & S &=& 1\,+\,5\,+\,7\,+\,11\,+\,13\,+\,17\,+\,19\,+ \,23\,+\,\cdots\,+\,295\,+\, 299 \;\;\;\;\;\;\;\;\\ \\ \\
& S &=& (1+5)\,+\,(7+11)\,+\,(13+17)\,+\,(19+23)\,+\, \cdots\,+\,(295+299) \;\;\;\;\;\; \\ \\ \\
& S &=& 6\,+\,18\,+\,30\,+\,42\,+\,54\,+\,66\,+\,\cdots\,+ \,594 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\; \\ \\ \\
& S &=& 6(1\,+\,3\,+\,5\,+\,7\,+\,9\,+\,\cdots\,+\,99)\;\; \;\;\;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;
\end{array}$

$\text{The sum of the fractions is: }\:F \;=\;\frac{S}{6} \;=\;1\,+\,3\,+\,5\,+\,\cdots\,+\,99$

The sum of the first 50 odd numbers is $50^2.$

Therefore: $\:F \:=\:2500.$

May 8th, 2015, 03:10 AM   #6
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Quote:
 Originally Posted by greg1313 $\displaystyle \sum_{k=1}^{n}(2k-1)=n^2$

Greg, could you please explain the procedure?

 May 8th, 2015, 05:09 AM #7 Global Moderator   Joined: Dec 2006 Posts: 20,469 Thanks: 2038 It's an arithmetic progression (A.P.) with first term 1 and last term n-1. 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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