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Math Help - Proof by Mathematical induction!

  1. #1
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    Proof by Mathematical induction!

    Hi there,

    I have this assignment, and I thought I was on the right path, but when I received it back with suggestions to fix it, I'm still not sure where to go from here. Thanks for your help!

    Create a proof by mathematical induction that demonstrates that the sum of the first n even numbers is equal to n(n + 1).

    Let P be the proposition that the sum of the first n even numbers is n(n+1);
    P(n) : 2 + 4 + 6+ + 2n = n(n+1)

    With mathematical induction we first need to show that this is true when n=1
    Since we are only discussing even numbers, we will start with 2:
    2=n(n+1); n=1 so; 2=1(1+1), following simple algebra we find that this is true for n=1:
    2=1(2) and 2=2

    Inductive Step: Assume that this is true for n=k; so
    P(k) : 2 + 4 + 6 + +2k =k(k+1)
    2+4+6+8+10+12+2(7)=7(7+1)
    56 = 56

    (HERE I WAS TOLD I NEED TO SOLVE FOR ALL N=K, not just 7....)


    Now we need to prove that this is also true for (k+1) whenever P(k) is true,
    2 + 4 + 6 + + 2k + 2(k+1) = (k+1)(k+2)
    2+4+6+8+10+12+2(7)+2(7+1) = 7^2 + 3(7) + 2
    56 + 16 = 49 + 21 +2
    72 = 72

    (Here I was told I was wrong again, and it was suggested that I add 2(k+1) to both sides, but I don't see how that works out..... help I'm stuck!)

    THanks.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Conorsmom View Post
    Hi there,

    I have this assignment, and I thought I was on the right path, but when I received it back with suggestions to fix it, I'm still not sure where to go from here. Thanks for your help!

    Create a proof by mathematical induction that demonstrates that the sum of the first n even numbers is equal to n(n + 1).

    Let P be the proposition that the sum of the first n even numbers is n(n+1);
    P(n) : 2 + 4 + 6+ … + 2n = n(n+1)

    With mathematical induction we first need to show that this is true when n=1
    Since we are only discussing even numbers, we will start with 2:
    2=n(n+1); n=1 so; 2=1(1+1), following simple algebra we find that this is true for n=1:
    2=1(2) and 2=2

    Inductive Step: Assume that this is true for n=k; so
    P(k) : 2 + 4 + 6 + … +2k =k(k+1)
    2+4+6+8+10+12+2(7)=7(7+1)
    56 = 56

    (HERE I WAS TOLD I NEED TO SOLVE FOR ALL N=K, not just 7....)


    Now we need to prove that this is also true for (k+1) whenever P(k) is true,
    2 + 4 + 6 + … + 2k + 2(k+1) = (k+1)(k+2)
    2+4+6+8+10+12+2(7)+2(7+1) = 7^2 + 3(7) + 2
    56 + 16 = 49 + 21 +2
    72 = 72

    (Here I was told I was wrong again, and it was suggested that I add 2(k+1) to both sides, but I don't see how that works out..... help I'm stuck!)

    THanks.
    You have shown that its true for n=1.

    Now, assume it holds for n=k. Then show its true for n=k+1.

    Since 2+4+6+8+...+2k=k(k+1), we need to show that 2+4+6+8+...+2k+2(k+1)=(k+1)(k+2)

    Note that \underbrace{2+4+6+8+...+2k}_{k(k+1)}+2(k+1)=(k+1)(  k+2)

    The equation becomes k(k+1)+2(k+1)=(k+1)(k+2).

    Expanding out the left side, we see that

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

    \implies k^2+k+2k+2=(k+1)(k+2)

    \implies k^2+3k+2=(k+1)(k+2)

    Factoring the left side of the equation, we now end up with (k+1)(k+2)=(k+1)(k+2)\blacktriangleleft.

    We have completed the proof by induction.

    Does this make sense?

    --Chris
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  3. #3
    Member
    Joined
    Oct 2008
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    Solution

    Alright, so you got to the inductive step okay.
    So you would have:
    P(k) = 2 + 4 + 6 + +2k =k(k+1)

    Now, you would have to show that it holds for P(k+1).
    P(k+1) = 2 + 4 + 6 + +2k + 2(k+1) = (k+1)*((k+1)+1)

    Simplify:
    P(k+1) = 2 + 4 + 6 + +2k + 2(k+1) = (k+1)*(k+2)

    Now for the left hand side, we know 2 + 4 + 6 + ... + 2k = k(k+1), so use substitution and you get:
    P(k+1) = k(k+1) + 2(k+1) = (k+1)*(k+2)

    Now, work with the left hand side:
    k^2 + k + 2k +1
    k^2 + 3k + 1

    Factor this to get:
    (k+1)*(k+2), which is exactly what the right side of the equation is. You have now shown that the property holds for P(k+1) and you are done. I hope this helps.
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