justforlaugh Posted July 15, 2009 Report Share Posted July 15, 2009 (edited) HEY,PLEASE HELP (:(1+ax)^n=1-12x+63x^2-....Find out the values of a and n.These are the 2 binomial equation i obtained before i got stuck.nC2(a)^2=63 and n(ax)=-12the answers are a=-3/2 and n=8THANKS SO MUCH Edited July 15, 2009 by justforlaugh Reply Link to post Share on other sites More sharing options...
sweetnsimple786 Posted July 15, 2009 Report Share Posted July 15, 2009 (edited) HEY,PLEASE HELP (:(1+ax)^n=1-12x+63x^2-....Find out the values of a and n.These are the 2 binomial equation i obtained before i got stuck.nC2(a)^2=63 and n(ax)=-12 "n(ax)=-12" This isn't right--you already canceled the x on the right side, but you didn't on the left side. So what you're left with is n(a)=-12nC2(a2)=63 --> n!/[(n-2)!2] * a2 = 63 --> n(n-1)(a2) / 2 = 63 --> n(n-1)(a2) = 126 **Edit: If you didn't get how to simplify the factorial, think of n as any number. For example, 12. so 12!/(12-2)! = 12*11*10!/(12-2)! = 12*11*10!/10! = 12*11. **Now it's a system of equations... just gotta substitute and solve.n(a)=-12 so a= -12/nNow plug into the second equationn(n-1)(144/n2)=126 Simplify(n-1)(144/n)=126(n-1)(144)=126n144n - 144=126n18n - 144=018n=144n=8Now for a.a=-12/n so a=-12/8 = -3/2If this is an MC question, plug in values. If not, guess some numbers if you have no clue. Edited July 15, 2009 by sweetnsimple786 Reply Link to post Share on other sites More sharing options...
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