一个等比数列前n项和为s,前n项的倒数的和为T,则前n项的积是多少?
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一个等比数列前n项和为s,前n项的倒数的和为T,则前n项的积是多少?
一个等比数列前n项和为s,前n项的倒数的和为T,则前n项的积是多少?
一个等比数列前n项和为s,前n项的倒数的和为T,则前n项的积是多少?
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)
S=a1(1-q^n)/(1-q) 得
q^n-1=(q-1)*S/a1
T=(1/a1)(1-(1/q)^n)/(1-(1/q))=(q^n-1)/[a1*q^(n-1)*(q-1)]
=(q-1)*S/[a1^2*q^(n-1)*(q-1)]
=S/[a1^2*q^(n-1)]
得
a1^2*q^(n-1)=S/T
a1*a2*a3*....
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S=a1(1-q^n)/(1-q) 得
q^n-1=(q-1)*S/a1
T=(1/a1)(1-(1/q)^n)/(1-(1/q))=(q^n-1)/[a1*q^(n-1)*(q-1)]
=(q-1)*S/[a1^2*q^(n-1)*(q-1)]
=S/[a1^2*q^(n-1)]
得
a1^2*q^(n-1)=S/T
a1*a2*a3*...*an=a1^n*q^[1+2+3+...+(n-1)]
=a1^n*q^[n(n-1)/2]
=[a1^2*q^(n-1)]^(n/2)
=(S/T)^(n/2)
收起
(s/t)的n/2次方