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Hard inequalities question

  1. Jun 23, 2008 #1
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Jun 23, 2008 #2

    HallsofIvy

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    If (f) is the part you are having trouble with, then presumably you have already proved that [itex]2x_n^2- (2n-1)x- (n+1)= 0[/itex] (part (e)). Now you want to find the smallest n such that [itex]x_n< n+ 0.05[/itex]. You could, for example, solve that using the quadrative formula and compare the solutions to n+ 0.05. Have you calculated some values of [itex]x_n[/itex]? What are [itex]x_0[/itex] [itex]x_1[/itex], etc.?
     
  4. Jun 23, 2008 #3
    Thanks for the help. Im still confused as to how the markscheme answers have come about which I attached above.

    Thanks
     
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