CSSA 2005 Q7 Solution (1 Viewer)

[Libbster]

Does anybody have the solutions for the binomial theorem q in q7 CSSA 2005. I was doing this paper as a practice paper and can't figure it out!

Thanks

 

[acmilan]

Mr Hong Kong answered this question before:

S=1+x+x²+...+x^(2n-1)
You can see that it is a GP, the ratio is x
so S=1(x^(2n)-1)\(x-1)

Given x²-x-1=0
then x²=1+x
so x^2n=(1+x)^n

the general term for (1+x)^n is: nCr x^r
so x^2n=(1+x)^n=sigma nCr x^r { from r=0 to r=n }
you can see that when r=0
then sigma nCr x^r=1
so x^2n-1=sigma nCr x^r { from r=1 to r=n }

now consider x²-x-1=0 again
x²-x=1
x(x-1)=1
(x-1)=1\x
.: 1\(x-1)=x
so now S=(x^(2n)-1)\(x-1)=(x)*sigma nCr x^r { from r=1 to r=n }
=sigma nCr x^(r+1) { from r=1 to r=n }

tink dats rite

 

[Libbster]

Oops, sorry, I didn't see the thread. Thanks!