Let Fn be the nth Fibonacci number, where as usual F1=F2=1 and Fn+1=Fn+Fn−1. Then k=2∏100(Fk−1Fk−Fk+1Fk)=FbFafor some positive integers a and b. Enter the ordered pair (a,b).
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We have that Fk−1Fk−Fk+1Fk=Fk−1Fk+1FkFk+1−FkFk+1Fk−1Fk=Fk−1Fk+1FkFk+1−Fk−1Fk=Fk−1Fk+1Fk(Fk+1−Fk−1)=Fk−1Fk+1Fk2.Thus, k=2∏100(Fk−1Fk−Fk+1Fk)=k=2∏100Fk−1Fk+1Fk2=F1⋅F3F22⋅F2⋅F4F32⋅F3⋅F5F42⋯F98⋅F100F992⋅F99⋅F101F1002=F1⋅F101F2⋅F100=F101F100.Therefore, (a,b)=(100,101).