
#1
Jun2308, 09:47 AM

P: 228





#2
Jun2308, 10:25 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,886

If (f) is the part you are having trouble with, then presumably you have already proved that [itex]2x_n^2 (2n1)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.?




#3
Jun2308, 01:20 PM

P: 228

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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