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Math Help - consecutive odd integers

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    consecutive odd integers

    The sum of 3 consecutive odd integers is k. In terms of k, what is the sum of the 2 smaller of these integers?

    Do I need to use substitution somehow?? (The answer given is \frac {2k}{3} - 2 but can't figure it out..)
    Last edited by mjoshua; March 28th 2010 at 02:15 PM.
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    Quote Originally Posted by mjoshua View Post
    The sum of 3 consecutive odd integers is k. In terms of k, what is the sum of the 2 smaller of these integers?

    Do I need to use substitution somehow?? (I know it's gonna be \frac {2k}{3} - 2 but can't figure it out..)
    k - (largest odd integer).

    If you post the whole question you might get an answer that's more suitable for your purpose.
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    Quote Originally Posted by mjoshua View Post
    The sum of 3 consecutive odd integers is k. In terms of k, what is the sum of the 2 smaller of these integers?

    Do I need to use substitution somehow?? (I know it's gonna be \frac {2k}{3} - 2 but can't figure it out..)
    Let n be the middle odd integer so we have

    k - n+2

    as Mr F said, you'll need to expand
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    That IS the entire question (it's taken out of my book as is)
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    Quote Originally Posted by mjoshua View Post
    The sum of 3 consecutive odd integers is k. In terms of k, what is the sum of the 2 smaller of these integers?

    Do I need to use substitution somehow?? (The answer given is \frac {2k}{3} - 2 but can't figure it out..)
    Perhaps the idea of an arithmetic sequence can help us.

    Let n be the middle number so that our sequence is (n-2)+n+(n+2) = k

    For 3 consecutive odd numbers the mean will be equal to the middle number. Hence

    n = \frac{k}{3} is the middle number.

    Hence the smaller number is: n-2 = \frac{k}{3} - 2.

    The question is asking us to find the sum of the two smaller numbers:

    (n-2)+n = \left(\frac{k}{3} - 2 \right) + \frac{k}{3} = \frac{2k}{3} - 2

    which is the book's answer
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    Quote Originally Posted by e^(i*pi) View Post
    Perhaps the idea of an arithmetic sequence can help us.

    Let n be the middle number so that our sequence is (n-2)+n+(n+2) = k
    Why can we use this sequence though? (Should it be a consecutive integer odd sequence? If you try n = 0, or n = 2 it's not completely odd.
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    Quote Originally Posted by mjoshua View Post
    Why can we use this sequence though? (Should it be a consecutive integer odd sequence? If you try n = 0, or n = 2 it's not completely odd.
    Neither 0 nor 2 are odd integers.

    If you think about an arithmetic sequence it is defined as there being a common difference between each term. For consecutive odd integers this is 2. n is 2 more than n-2 and n+2 is 2 more than n.

    If we try n=3 and hence k = 1+3+5 = 9

    (3-2)+3 = \frac{2(9)}{3} - 2 and sure enough 4=4
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    But instead of declaring "n" just an odd integer, shouldn't we use a sequence like 2n + 1 so that no matter what you put it for n (even or odd) you still get an odd integer? (so it's more inclusive?)
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    Quote Originally Posted by mjoshua View Post
    But instead of declaring "n" just an odd integer, shouldn't we use a sequence like 2n + 1 so that no matter what you put it for n (even or odd) you still get an odd integer? (so it's more inclusive?)
    You have been given the solution on a platter in post #5. Now it's your job to take some ownership of that solution and modify it if you want to express the odd numbers in the way you describe.
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    sum of odd integers

    Quote Originally Posted by mjoshua View Post
    But instead of declaring "n" just an odd integer, shouldn't we use a sequence like 2n + 1 so that no matter what you put it for n (even or odd) you still get an odd integer? (so it's more inclusive?)
    Hello Mjoshua,
    I was going to reply to this post yesterday but decided that eipi had adequately answered it I passed.Today you appear confused.Here is the way I would have replied.

    Three cons odd integers are n, n+2, n+4, the sum of these =k
    Solution of this equation n=k-6/3. The sum of the first two is 2n+2 and thisbecomes 2/3k-2 when you insert the value of n in terms of k.

    bjh
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    Quote Originally Posted by mr fantastic View Post
    You have been given the solution on a platter in post #5. Now it's your job to take some ownership of that solution and modify it if you want to express the odd numbers in the way you describe.
    And I can see that solution. I was just asking a simple question, if was to better to instead create a "proof" for n as any positive or negative integer that covers all cases. I guess you missed that..
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