一个等比数列的前n项和为S，前n项的倒数和为T，求此数列的前n项之和

a(n) = aq^(n-1), aq 不等于0.
1/a(n) = 1/aq^(1-n) = (1/a)(1/q)^(n-1).

M(n) = a(1)*a(2)*...*a(n) = a^nq^[1+2+...+(n-1)] = a^nq^[n(n-1)/2]

若q = 1, 则，T = n/a,因此，T不等于0.a = n/T,
S = na = n*n/T = n^2/T, n = (S*T)^(1/2).
a = n/T = (S*T)^(1/2)/T = (S/T)^(1/2).
M(n) = a^n = (S/T)^(n/2)， n = 1,2,...

若q不等于1，则S = a[q^n - 1]/[q - 1], 1/a = [q^n - 1]/[S(q-1)].
T = (1/a)[(1/q)^n - 1]/(1/q - 1) = [q^n - 1][1/q^n - 1]/[S(q-1)(1/q - 1)],

T[S(q-1)(1/q - 1)] = [q^n - 1][1/q^n - 1],

q^(n-1)TS(q-1)^2 = [q^n - 1]^2,

q^[(n-1)/2](TS)^(1/2) = [q^n - 1]/(q-1),
aq^[(n-1)/2](TS)^(1/2) = a[q^n - 1]/(q-1) = S,
aq^[(n-1)/2] = (S/T)^(1/2)
M(n) = a^nq^[n(n-1)/2] = {aq^[(n-1)/2]}^n = [(S/T)^(1/2)]^n = (S/T)^(n/2)

一个等比数列的前n项和为S，前n项的倒数和为T，求此数列的前n项之和
一个等比数列的前n项和为S ？