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Math Help - Induction

  1. #1
    CPR
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    Induction

    Okay I must prove


    Base n=1, K=1
    1+4+7......+(3n-2) = n(2n-1)

    (3k-2) = k(2k-1)
    (3k-2)+(k+1) = (k+1)(2k+1-1)
    = (k+1)(2k+2-1)
    = (k+1)(2k+1)

    now I am lost, please show me what to do!
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  2. #2
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    Hello, CPR!

    Where are you getting these problems?

    Most of them are simply not true . . .


    1+4+7 + \hdots + (3n-2) \:= \:n(2n-1)

    Did you try the formula at all?


    For n = 5, the left side is: .  1 + 4 + 7 + 10 + 13 \;=\;35

    The right side is: . 5(2\!\cdot\!5-1) \;=\;45

    So they are lying to us . . .

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  3. #3
    CPR
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    We had to make up problems then solve using mathematical induction. I guess I wasnt doing a good job making them up.


    This is one I have that should work

    1+5+9.......+(4n-3)= n(3n-2)

    Okay,
    Base is n=1, k+1=1

    1+5+9.....+(4n-3)+(k+1)= (k+1)(3k+1-2)
    = (K+1)(3k-1)

    Okay, please show me what to do now,for I'm lost I know I must make LHS equal RHS, how do I do this? Thanks
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  4. #4
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    Well, CPR, it's not that hard to make problems like that, if you use the formula of arithmetic progress (which Gauss invented at the age of 10!):
    a_{1}+a_{2}+...+a_{n}= \frac{a_{1}+a_{n}}{2}n
    Now, you can set up the LAST factor of your sum, let's say (4n-3), which is at your problem. For n=1, it's 4x1-3=1, for n=2 it's 4x2-3=5 and so on, therefore your sum indeed is 1+5+9+...+(4n-3). Now, the right side of your problem will be found by using the Gauss' formula. Hence must be \frac{1+(4n-3)}{2}n=(2n-1)n.

    So, now, you know your -correct- right side of your problem, and you can apply induction.

    Take a look at some similar problems:
    1. 1+7+13+...+(6n-5) = n(3n-2)
    2. 5+7+9+...+(2n+3)= n(n+2)
    3. 4017+6025+...+(2008n+2009)=n(1004n+3013)



    PS I hope i didn't make any mistake on the operations :-).
    Last edited by Gauss; February 28th 2008 at 10:33 AM.
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  5. #5
    CPR
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    Thanks, but I still do not know what to do. Why do I need to divide by 2? Why has n(3n-2) become (2n-1)? As you can see I'm totally confused on your explanation.
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  6. #6
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    Hello, CPR!

    You must come up with a statement that is true.


    1 + 5 + 9 + \hdots + (4n-3) \:= \:n(3n-2) . . . . not true!

    Do you ever check your formulas?
    This one isn't true either . . .

    For n = 4, the left side is: .  1 + 5 + 9 + 13 \:=\:28

    But the right side is: . 4(3\!\cdot\!4-2) \;=\;40


    Here's the correct formula:

    1 + 5 + 9 + \hdots + (4n-3) \;=\;n(2n-1)


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    And you evidently don't understand the induction process.

    When we write the S(k+1) statement,
    . . it does not mean "stick in a plus-one somewhere."

    It means: Replace k with k+1 . . . there's a big difference.

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  7. #7
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    As you were shown, the series is 1+5+9+....+(4n-3)=n(2n-1)

    Check for n=1:

    1(2(1)-1)=1..........true.

    Assume true for P_{k}. Therefore,

    1+5+9+........+(4k-3)=k(2k-1)

    We must show it is true for P_{k+1}

    That is, we have to show that 1+5+9+...+(4(k+1)-3)=(k+1)(2(k+1)-1)

    Let's add 4k+1 to both sides and see what happens:

    1+5+9+......+(4k-3)+(4k+1)=k(2k-1)+(4k+1)

    The right side: k(2k-1)+(4k+1)=2k^{2}+3k+1=(k+1)(2k+1)=\boxed{(k+1)(2(k  +1)-1)}

    This shows that P_{k+1} is true and the proof is complete.

    See how the k in our previous series is replaced with a k+1?. I stepped through it hoping you can see it and can do another. You can't just throw it in. It has to be shown.
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  8. #8
    CPR
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    Thank you so very much Galactus! For taking the time to explain it and not be critical. I am sorry that the formulas were not correct. But you corrected it and showed me how to do it. That is what I needed. I now see it and understand it and the process. Again thanks! I got it!
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