Question

高二数列

Answer 1

a(n) = aq^(n-1),
27 = a(1)a(2)a(3) = a*aq*aq^2 = a^3q^3 = (aq)^3, aq = 3.
若q = 1, 则 a(n) = a = 3.
s(n) = na = 3n.
s(2n) = 3(2n)= 6n.
a(2n-1) = a = 3,
a(1)+a(3)+...+a(2n-1) = 3n.
s(2n) = 4[a(1)+a(3)+...+a(2n-1)] = 4(3n) = 12n, 与s(2n) = 6n矛盾。
因此，q不为1。

s(n) = a[q^n - 1]/(q-1).
s(2n) = a[q^(2n) - 1]/(q-1).

a(2n-1) = aq^(2n-1-1)= aq^(2n-2) = a(q^2)^(n-1),
4[a(1)+a(3)+...+a(2n-1)] = 4a[(q^2)^n - 1]/(q^2-1) = s(2n) = a[q^(2n)-1]/(q-1),

4[q^(2n)-1]/[(q-1)(q+1)] = [q^(2n) - 1]/(q-1),
4/(q+1) = 1,
q+1 = 4,
q = 3.
a = 3/q = 1.

a(n) = 3^(n-1).
a(6) = 3^(6-1) = 3^5 = 3*81 = 243