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Math Help - Some help with simple "how to prove thing"

  1. #1
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    Some help with simple "how to prove thing"

    Hello everyone!
    to start with I am new to this forum and I come from sweden so I am not that good in english mathematic terms. But I will try my best. Also I am going in highschool and not college/university but sinse the course I am taking is named "Descrete mathe" I thought it would fit quiet good here. Anyway, I think this is very easy but I am in year 1 in highschool so for me it's hard.

    I need to prove that n = 1,2,3 ...
    is working for
    a) 2 + 4 +...+ 2n = n(n+1)
    b) 1 + 2 + 4 +...+ 2^n = 2^(n+1)-1

    I cant even figure out the A..
    This is how I do it, please help me !
    n=1 VL=2, HL=1(1+1)=2, VL=HL

    n=p,
    2 + 4 +...+ 2p = p(p+1)

    n=p+1,
    2 + 4 +...+ 2p + (2p+1) = (p+1)(p+2)
    p(p+1)+(2p+1) = (p+1)(p+2)
    p^2+p+2p+1 = p^2+p+2p+2

    Now why is that wrong, I am doing something wrong..
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  2. #2
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    Nevermind I did discovere the problem with some help, i have put 2p inside the (2p+1) when it should be 2(p+1). Now it is just b) left!
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  3. #3
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    Hello, Mquis!

    Welcome aboard!

    It looks like you're using Inductive Proofs.


    a) Prove: . 2 + 4 + 6 + \cdots...+ 2n \:=\:n(n+1)

    Verify S(1)\!:\;\;2 \:=\:1(1+1) \quad\Rightarrow\quad 2 \:=\:2 . . . true!

    Assume S(k)\!:\;\;2 + 4 + 6 + \cdots + 2k \:=\:k(k+1)

    Add 2(k+1) to both sides: . 2 + 4 + 6 + \cdots + 2k + 2(k+1) \:=\:k(k+1) + 2(k+1)

    . . The right side is: . k^2 + k + 2k + 2 \:=\:k^2+3k + 2 \:=\:(k+1)(k+2)

    The equation becomes: . 2 + 4 + 6 + \cdots + 2(k+1) \:=\:(k+1)(k+2)

    This is S(k+1) . . . The inductive proof is complete.




    b) Prove: . 1 + 2 + 4 + \cdots + 2^n \:= \:2^{n+1}-1

    Verify S(1)\!:\;\;1 +2\;=\;2^2-1\quad\Rightarrow\quad 3 \:=\:3 . . . true!

    Assume S(k)\!:\;\;1 + 2 + 4 + \cdots + 2^k \:=\:2^{k+1} - 1

    Add 2^{k+1} to both sides: . 1 + 2 + 4 + \cdots + 2^k + 2^{k+1} \;=\;2^{k+1}-1 + 2^{k+1}

    . . The right side is: . 2\cdot2^{k+1} - 1 \;=\;2^{k+2}-1

    The equation becomes: . 1 + 2 + 4 + \cdots + 2^{k+1} \;=\;2^{k+2}-1

    This is S(k+1) . . . The inductive proof is complete.



    Edit: I see that you figured it out . . . Good for you!
    .
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  4. #4
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    I just wonder 2 things.


    1.How can you say that you have prooved it... dont the sides have to be exactly? Even if that is understood.

    2.How do you write codes like that?

    Edit; 3. Did I post this in the right section?
    Last edited by Mquis; January 15th 2008 at 07:57 AM.
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