Enter the recursive rule for the geometric sequence.
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1 answer:
Explicit rule: a(n)=(2/5)(5^(n-1)) For a recursive rule, we need to express a(n) in terms of a(n-1), which we can obtain from the explicit rule a(n)=(2/5)(5^(n-1)) substitute n-1 for n above a(n-1)=(2/5)(5^((n-1)-1)) =(2/5)(5^n-2) Divide: a(n)/a(n-1)=(2/5)(5^(n-1)) / ((2/5)(5^(n-2))) =1/5^(-1) =5 Therefore, multiplying both sides by a(n-1) a(n)=5 a(n-1) a(1)=(2/5)(5^(1-1))=2/5 So the recursive rule isa(1)=2/5, a(n)=5 a(n-1)
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