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May 6th, 2015, 06:01 PM   #1
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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.
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May 6th, 2015, 06:30 PM   #2
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$\displaystyle \sum_{k=1}^{n}(2k-1)=n^2$
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Last edited by greg1313; May 6th, 2015 at 08:26 PM.
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May 6th, 2015, 07:47 PM   #3
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I think you intended $\displaystyle \sum_{k=1}^{n}(2k-1)=n^2\!$.
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May 6th, 2015, 08:30 PM   #4
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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).
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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.





The sum of the first 50 odd numbers is

Therefore:

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

Greg, could you please explain the procedure?
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May 8th, 2015, 05:09 AM   #7
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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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