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Two variabe induction question

  1. Nov 28, 2008 #1
    n1 is the smallest whole number for which this inequality works :

    (1+x)^n >1 +n*x + n*x^2

    also i am given that x>0

    find n1

    and prove this inequality for every n=>n1 by induction.

    the base case:

    (1+x)^n1 >1+n1*x + n1*x^2

    i think its correct because i was told that this inequality works for n1.

    n=k step we presume that this equation is true :

    equation 1: (1+x)^k >1 +k*x + k*x^2

    n=k+1 step we need to prove this equation:
    equation 2: (1+x)^(k+1) >1 +(k+1)*x + (k+1)*x^2

    now i need to multiply equation1 by sum thing
    if a<b<c
    then a<c

    how to do this thing in this case?
    Last edited: Nov 28, 2008
  2. jcsd
  3. Nov 28, 2008 #2
    i multiplied equation 1 by (1+x)
    (1+x)^(k+1) >(1 +k*x + k*x^2) (1+x)

    i tried to do this
    but its not working
    (1+x)^(k+1) >(1 +k*x + k*x^2) (1+x)>(1 +k*x + k*x^2)

    (1+x)^(k+1) >(1 +k*x + k*x^2)

  4. Nov 28, 2008 #3
    First, have you found n1? I suggest using the strong form of induction: Assume the inequality is valid for all whole numbers less n.
  5. Nov 28, 2008 #4
    what is the general way of finding n1?

    whats strong form of induction?

    i know

  6. Nov 28, 2008 #5
    I don't know of a general way. I would try n1 = 1, 2, 3, etc. and see which one works.

    Instead of assuming n = k and proving n = k + 1, you assume the proposition is true for all k < n and then prove the proposition for n.
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