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Math Help - Find the least value of n in an arithmetic sequence

  1. #1
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    Find the least value of n in an arithmetic sequence

    Find the least value of n for which \sum_{r=1}^n (4r-3) > 2000

    How do I do this?
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    Quote Originally Posted by Viral View Post
    Find the least value of n for which \sum_{r=1}^n (4r-3) > 2000

    How do I do this?
    Furst thing you do is simplify it down by extracting the constants outside:

    \sum_{r=1}^n (4r-3) = 4 \sum_{r=1}^n r - \sum_{r=1}^n 3 = 4 \sum_{r=1}^n r - 3n.

    Then you know (or ought to know, or be able to find out, or work out) what  \sum_{r=1}^n r is.

    Then you have an inequality in n to solve. Good luck.
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    Hmm, I'm not really sure what you did there =\ . We've never covered that.
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    Quote Originally Posted by Viral View Post
    Find the least value of n for which \sum_{r=1}^n (4r-3) > 2000

    How do I do this?
    \sum_{r=1}^n (4r-3) = 4 \left( \sum_{r=1}^n r\right) - 3n and you should know the formula to substitute for \sum_{r=1}^n r. Then solve for smallest positive integer value of n such that it's greater than 2000.

    If you need more help please show all your working and say where you get stuck.
    Last edited by mr fantastic; October 17th 2009 at 02:53 PM. Reason: Fixed a bit of latex
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    Senior Member MacstersUndead's Avatar
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    Quote Originally Posted by Viral View Post
    Hmm, I'm not really sure what you did there =\ . We've never covered that.
    Sigma sums preserve multiplication by a constant and preserve addition. You can prove this using the definition of sigma sums.
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    Is \sum_{r=1}^n(4r-3) = 4(\sum_{r=1}^n(r-3))?
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    Senior Member MacstersUndead's Avatar
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    Quote Originally Posted by Viral View Post
    Is \sum_{r=1}^n(4r-3) = 4(\sum_{r=1}^n(r-3))?
    No, but \sum_{r=1}^n(4r-3) = 4(\sum_{r=1}^n(r-(3/4))
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    Hmm, I kind of understand that. Are there other ways to solve this as our teacher specifically said we should be able to solve all questions with what we have been taught.
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    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by Viral View Post
    Hmm, I kind of understand that. Are there other ways to solve this as our teacher specifically said we should be able to solve all questions with what we have been taught.
    Fair comment. Can you document on this thread everything you've been taught? (joke)
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    To be honest, I would but I don't know how I'd explain it. We've learned how to calculate all terms in a sequence, how to find a, d, n and l. And this is our first look at the summation notation. If there are other methods, and we've been taught it, it should hit me when I see it.
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    Quote Originally Posted by Viral View Post
    Hmm, I kind of understand that. Are there other ways to solve this as our teacher specifically said we should be able to solve all questions with what we have been taught.
    The first term in the sum is 4 - 3 = 1. Therefore a = 1.

    The common difference between each term is 4. Therefore d = 4.

    There are n terms.

    Substitute all that into your formula for the sum of an arithmetic series and then do with the result what I said to do in my first reply.
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    S_n = \frac{n}{2}[2a + (n-1)d]

    Just to check, is this the formula for calculating the sum?
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    Quote Originally Posted by Viral View Post
    S_n = \frac{n}{2}[2a + (n-1)d]

    Just to check, is this the formula for calculating the sum?
    It's just as easy for you to go to a textbook or use Google as it is for any of us ....
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    It's a help forum after all >> . My textbook has been misplaced, hence why I'm here. I'm also here because google brought me here.
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    Quote Originally Posted by Viral View Post
    It's a help forum after all >> . My textbook has been misplaced, hence why I'm here. I'm also here because google brought me here.
    Yes, help as in help you to help yourself. Really, why should we be the ones that check a formula using Google when you can easily do it yourself? That's your job, not our job.
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