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Math Help - Can someone guide me with this simple induction problem?

  1. #1
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    Can someone guide me with this simple induction problem?

    Prove: 2 + 4 + 6 + ... + 2n= n^2+n for all integers n >= 1.

    Proof:

    P(n):2+4+6+...+2n= n^2+n

    P(2):2+4+6+...+2n= 2^2+2= 6 which is true.

    Suppose that P(k): 2+4+6+...+2k= k^2+k where k>=1.

    2+4+6+...+2k+(2k+1)= (2k+1)^2+(2k+1)= 4k^2+2k+2k+1 +(2k + 1)= 4k^2+6k+2= 2(2k^2+3k+2)= This is as far as I have gotten

    Can someone please help? I am trying to prepare for a Final Exam

    Thanks
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  2. #2
    Super Member Deadstar's Avatar
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    Quote Originally Posted by matthayzon89 View Post
    Prove: 2 + 4 + 6 + ... + 2n= n^2+n for all integers n >= 1.

    Proof:

    P(n):2+4+6+...+2n= n^2+n

    P(2):2+4+6+...+2n= 2^2+2= 6 which is true.

    Suppose that P(k): 2+4+6+...+2k= k^2+k where k>=1.

    2+4+6+...+2k+(2k+1)= (2k+1)^2+(2k+1)= 4k^2+2k+2k+1 +(2k + 1)= 4k^2+6k+2= 2(2k^2+3k+2)= This is as far as I have gotten

    Can someone please help? I am trying to prepare for a Final Exam

    Thanks
    Assume true for P(k)

    Then P(k+1) = P(k) + 2(k+1) = k^2 + k + 2k + 2.

    = k^2 + 2k + 1 + (k+1) = (k+1)^2 + (k+1).

    In general with these induction things usually the k+1 step involves saying P(k+1) = P(k) + ... Then sub in what P(k) equals (not the sequence though, in this case P(k) = 2^k + k).
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  3. #3
    baz
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    Quote Originally Posted by matthayzon89 View Post
    Prove: 2 + 4 + 6 + ... + 2n= n^2+n for all integers n >= 1.

    Proof:

    P(n):2+4+6+...+2n= n^2+n

    P(2):2+4+6+...+2n= 2^2+2= 6 which is true.

    Suppose that P(k): 2+4+6+...+2k= k^2+k where k>=1.

    2+4+6+...+2k+(2k+1)= (2k+1)^2+(2k+1)= 4k^2+2k+2k+1 +(2k + 1)= 4k^2+6k+2= 2(2k^2+3k+2)= This is as far as I have gotten

    Can someone please help? I am trying to prepare for a Final Exam

    Thanks
    you made a mistake in inductive step,so yo were stuck. it 2(k+1) rather than (2k+1).you should have added 2(k+1) on bothsides.
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