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Math Help - Just checking my answers.

  1. #1
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    Just checking my answers.

    a1 = 3, ak+1 = ak − 2

    What is the value of the 4th term? (Select one.)

    -9
    -5
    -3
    1
    43

    I selected -3 for this. I am pretty sure about this.



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

    Which expression below represents the given function? (Select one.)

    F(n) = n(n + 1)

    F(n) = (n − 1)(n + 1)

    F(n) = n(n − 1)

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

    F(n) = n2 + 2n

    I guessed at answer 4, so could do with some help if I got it wrong.

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  2. #2
    Super Member Aryth's Avatar
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    Could you explain the function:

    F(n) = n2 + 2n

    Is it 2n + 2n Or:

    F(n) = n^2 + 2n

    Or something else?
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  3. #3
    Moo
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    A Cute Angle Moo's Avatar
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    P(I'm here)=1/3, P(I'm there)=t+1/3
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    Hello,

    Quote Originally Posted by Tally View Post
    a1 = 3, ak+1 = ak − 2

    What is the value of the 4th term? (Select one.)

    -9
    -5
    -3
    1
    43

    I selected -3 for this. I am pretty sure about this.


    Ok for this.

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

    Which expression below represents the given function? (Select one.)

    F(n) = n(n + 1)

    F(n) = (n − 1)(n + 1)

    F(n) = n(n − 1)

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

    F(n) = n2 + 2n

    I guessed at answer 4, so could do with some help if I got it wrong.

    No it's not.

    Note that your sequence is 2(1+2+\dots +n)

    What's in brackets is equal to \frac{n(n+1)}{2} as sum of the first n consecutive numbers.

    Hence F(n)=2 \cdot \frac{n(n+1)}{2}=n(n+1)
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  4. #4
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    Thank you for your reply. I am still having trouble with Induction. For some reason this is a long old road.

    Could you explain why where the /2 came from as I don't see it in the function.
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  5. #5
    Super Member Aryth's Avatar
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    \sum_{i=1}^n 2i = 2\sum_{i=1}^n i

    Where:

    \sum_{i=1}^n i = \frac{n(n + 1)}{2}

    So:

    2\frac{n(n + 1)}{2} = n(n + 1)

    We have to verify it for the first case:

    P(1) = 1(1 + 1) = 2

    Now we make our assumption:

    P(k) = 2 + 4 + 6 + ... + 2k = k(k + 1)

    Now we have to show that:

    P(k + 1) = 2 + 4 + 6 + ... + 2k + 2(k + 1) = (k + 1)((k + 1) + 1)

     = P(k) + 2(k + 1)

     = k(k + 1) + 2(k + 1)

     = k^2 + k + 2k + 2

     = k^2 + 3k + 2

     = (k + 1)(k + 2)

     = (k + 1)((k + 1) + 1)

    Therefore the form n(n + 1) is valid for all n \in \mathbb{N}
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