已知{an}是各项均为正数的等比数列,且a3+a5=20,a4a6=256 求an通项公式
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a(n) = aq^(n-1),a>0,q>0.
20 = a(3)+a(5) = aq^2 + aq^4,
256 = a(4)a(6) = [aq^3][aq^5] = (aq^4)^2 = (16)^2,
aq^4 = 16,
aq^2 = 20 - aq^4 = 20-16=4,
q^2 = (aq^4)/(aq^2) = 16/4 = 4,q=2.
a = 4/q^2 = 4/4 = 1.
a(n) = 2^(n-1)
20 = a(3)+a(5) = aq^2 + aq^4,
256 = a(4)a(6) = [aq^3][aq^5] = (aq^4)^2 = (16)^2,
aq^4 = 16,
aq^2 = 20 - aq^4 = 20-16=4,
q^2 = (aq^4)/(aq^2) = 16/4 = 4,q=2.
a = 4/q^2 = 4/4 = 1.
a(n) = 2^(n-1)
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