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Mathematical induction problem

  1. Jun 16, 2012 #1
    Hello! First of all I have like 5 exercises I don't quite understand so will it be a problem if I create 5 new topics in the next 24h?
    1. The problem statement, all variables and given/known data
    Prove, by using mathematical induction that if [itex]x+1 \geq 0[/itex] then [itex](1+x)^n \geq 1+nx[/itex].

    2. Relevant equations



    3. The attempt at a solution
    Basic step: If [itex]n=1[/itex] then [itex]1+x \geq 1+x[/itex] which is true.
    Induction Step: Now making [itex]n=k[/itex] we get [itex](1+x)^k \geq 1+kx[/itex]. If the hypothesis holds for [itex]n=k[/itex] then it will hold for [itex]n=k+1[/itex]. Making [itex]n=k+1[/itex] we get [itex](1+x)^k + (1+x)^{k+1} \geq 1+(k+1)x[/itex]. And from this:
    [itex](1+x)^k(1+(1+x)) \geq 1+(k+1)x[/itex]
    But by our induction hypothesis [itex](1+x)^k \geq 1+kx[/itex] which means that [itex](1+x)^k(1+(1+x)) \geq 1+(k+1)x[/itex] is true.

    Thanks!
     
  2. jcsd
  3. Jun 16, 2012 #2

    SammyS

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    The induction step is:
    Assume the following is true: [itex](1+x)^k \geq 1+kx\,.[/itex]

    From that you need to show that the following is true: [itex] (1+x)^{k+1} \geq 1+(k+1)x\,.[/itex]​

    It looks to me as if you're assuming the hypothesis holds for n = k+1 .
     
  4. Jun 16, 2012 #3
    That typo is actually not a typo. And I'm trying to say that if the ##n=k## holds then I'll try to prove that ##n=k+1## also holds by doing ##(1+x)^k + (1^x)^{k+1}##. I can write it clearer in my language. But the main problem here is ##(1+x)^k + (1+x)^{k+1}##. Is this correct?
     
  5. Jun 16, 2012 #4

    SammyS

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    No, the following is not correct.
    If you substitute k+1 for n in [itex]\displaystyle (1+x)^n \geq 1+nx\,,[/itex] then you get [itex]\displaystyle (1+x)^{k+1} \geq 1+(k+1)x\,.[/itex]

    By the Way: [itex]\displaystyle (1+x)^{k+1}=(1+x)^k\cdot(1+x)\,,[/itex] it's not the same as [itex](1+x)^k+(1+x)^{k+1}\,.[/itex]

    Maybe you're thinking of [itex]\displaystyle \sum_{n=0}^{k+1}(1+x)^n=\left(\sum_{n=0}^{k}(1+x)^n\right)+(1+x)^{k+1}\,.[/itex]
     
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