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Solve this mathematical induction

  1. Aug 27, 2014 #1
    Can you help me with this exercise?

    [itex]1^{1}[/itex]+[itex]2^{2}[/itex]+[itex]3^{3}[/itex]+[itex]4^{4}[/itex]+...+[itex]n^{n}[/itex] = [itex]n^{n+1}[/itex]

    Thanks!

    PD. I was trying to solve, and i have this:

    [itex]1^{1}[/itex]=[itex]1^{1+1}[/itex] =
    1 = 1

    a) [itex]k^{k+1}[/itex]
    b) [itex]k+1^{k+1}[/itex] = [itex]k+1^{(k+1)+1}[/itex]

    a in b) [itex]k^{k+1}[/itex] + [itex]k+1^{k+1}[/itex] = [itex]k+1^{(k+1)+1}[/itex]

    ([itex]k)^{k}[/itex]([itex]k)^{1}[/itex]+([itex]k+1)^{k}[/itex]([itex]k+1)^{1}[/itex]=([itex]k+1)^{k}[/itex]([itex]k+1)^{1}[/itex]([itex]k+1)^{1}[/itex]

    I´m lost in this step . . . Thanks again!
     
    Last edited: Aug 27, 2014
  2. jcsd
  3. Aug 27, 2014 #2

    dextercioby

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    Science Advisor
    Homework Helper

    Where did you pick the exercise from? It looks wrong.
     
  4. Aug 27, 2014 #3
    The Algebra Teacher gives the exercise, but i dont know, maybe don´t have a solution . . . ?
     
  5. Aug 27, 2014 #4

    jedishrfu

    Staff: Mentor

    The series and its sum don't seem correct.

    Can you check your book again?

    1+2^2 =/= 2^(2+1)

    1+4 = 5 but 2^(2+1) = 8
     
  6. Aug 27, 2014 #5

    jedishrfu

    Staff: Mentor

    There are many series that look vaguely similar like

    1 + 2 + 3 + ... = (n*(n+1))/2
     
  7. Aug 27, 2014 #6
    A attach the picture of the book. . .
     

    Attached Files:

  8. Aug 27, 2014 #7

    jedishrfu

    Staff: Mentor

    Can you say there is no solution and prove that by example?
     
  9. Aug 27, 2014 #8
    The teacher said that exist a solution, but with the opinions and the examples i think that he´s lying. . . =o!
     
  10. Aug 27, 2014 #9

    jedishrfu

    Staff: Mentor

    Maybe strategically misdirecting you in order to have you discover this alternative...

    You could PM a PF Mentor to see if they can add some new insight...
     
  11. Aug 27, 2014 #10
    I am curious. What are the instructions for the exercise? You did not include them in the original post, and your picture does not show them either.
     
  12. Aug 28, 2014 #11
    Check the Attachment, the exercise says: "Demuestre por Induccion Matematica las siguientes preposiciones dadas" or in English, "Demonstrate by Mathematical Induction the following prepositions" . . . I think is a exercise that dont have a real solution, maybe the teacher´s proving us, . . . But I´ll need a good explanation for this, =)
     

    Attached Files:

  13. Aug 28, 2014 #12

    vanhees71

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    Science Advisor
    2016 Award

    The formula is obviously wrong.

    Much more interesting is the question, how to figure out such sums. One clever trick is to use an easier sum to derive the one you want to know. Here the obvious idea is to use the generating function
    [tex]g(x)=\sum_{k=0}^n \exp(k x).[/tex]
    This is a geometric series. You should prove, e.g., by induction, that it gives
    [tex]g(x)=\frac{\exp[(n+1) x)-1}{\exp x-1}.[/tex]
    Then you can get other sums by taking derivatives of this, e.g.,
    [tex]g'(x)=\sum_{k=0}^{n} k \exp(k x) \; \Rightarrow \; g'(0)=\sum_{k=0}^n k,[/tex]
    and so on. There's, of course a subtlety with taking [itex]x=0[/itex], which you also should discuss.
     
  14. Aug 28, 2014 #13

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, this is true for n= 1. But surely you saw that when n= 2, the left side is [itex]1^1+ 2^2= 1+ 4= 5[/itex] while the right side is [itex]2^3= 8[/itex].


     
  15. Aug 28, 2014 #14
    I´ve 1 curious question:

    If i wanna know what´s the correct serie for [itex]n^{n+1}[/itex], what´s the the procedure to follow?
     
  16. Aug 29, 2014 #15
    Use a Taylor series.
     
  17. Aug 29, 2014 #16

    jedishrfu

    Staff: Mentor

    Doing the first nine iterations:

    Code (Text):

    n=1:    1               sum/psum = 1
    n=2:    5               sum/psum = 5
    n=3:    32              sum/psum = 6
    n=4:    288             sum/psum = 9
    n=5:    3413                sum/psum = 11
    n=6:    50069               sum/psum = 14
    n=7:    873612              sum/psum = 17
    n=8:    17650828        sum/psum = 20
    n=9:    405071317       sum/psum = 22
     
    Processing IDE code used:

    Code (Text):

    void setup() {
     
     int nterms = 10;
     int psum=1, sum=0;
     
     for(int n=1; n<nterms; n++) {
        sum+=Math.pow(n,n);
        int iratio = sum/psum;
        println("n="+n+":\t"+sum+"\t\tsum/psum = "+iratio);
        psum=sum;
     }
     
    }
     
     
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