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Math Help - Immediate help needed! Induction and Recursion!

  1. #1
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    Exclamation Immediate help needed! Induction and Recursion!

    Okay, here's a question, I don't think the right answer is there!

    F(n)=2+4+6+...+2n

    Which expression represents the given function?

    (These are the options)

    F(n)=n(n+1)

    F(n)=n(n-1)

    F(n)=2n(2n+1)

    F(n)=(n-1)(n+1)

    F(n)=n^2+2n

    I thought at first that it was the last option, but I can't work it out just right..... I've now spent 2 hours on this question! and I need to go to bed soon! Thank you!
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  2. #2
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    Immediate help needed! Induction and Recursion!

    F(n)=2+4+6+...+2n is the same as F(n)=2(1+2+3+...+n)
    The sum of the numbers from 1 to n for positive n = n*(n+1)/2.

    So by substitution, F(n) = 2*n*(n+1) / 2 = n*(n+1), since the 2's cancel. This appears to be the first option in your list.

    - Steve J
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  3. #3
    Newbie momi's Avatar
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    Quote Originally Posted by Conorsmom View Post
    Okay, here's a question, I don't think the right answer is there!

    F(n)=2+4+6+...+2n

    Which expression represents the given function?

    (These are the options)

    F(n)=n(n+1)

    F(n)=n(n-1)

    F(n)=2n(2n+1)

    F(n)=(n-1)(n+1)

    F(n)=n^2+2n

    I thought at first that it was the last option, but I can't work it out just right..... I've now spent 2 hours on this question! and I need to go to bed soon! Thank you!
    To find the sum of terms in an arithmetic series, use the following formula.
    In this formula:
    Sn is the sum of the first n terms in a series
    n is the number of terms you are adding up
    a1 is the first term of the series
    an is the nth term of the series
    Here a1=2,n=n,an=2n,so the required sum F(n)is
    F(n)=n(2+2n)/2=2n(n+1)/2=n(n+1) which is the first one in your list.

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