若等比数列｛an｝的前n项和与积分别为S和T,数列｛1/an｝的前n项和为S',求证T^2=(S/S')^n

Question

若等比数列｛an｝的前n项和与积分别为S和T,数列｛1/an｝的前n项和为S',求证T^2=(S/S')^n

Answer 1

an = a1 * q^(n-1)
S = a1 * (q^n -1)/(q-1)
T = a1 * a2 * a3 …… = a1^n * q^[1 + 2 + …… + (n-1)]
= a1^n * q^[n(n-1)/2]
T^2 = a1^(2n) * q^[n(n-1)]
1/an = (1/a1) * (1/q)^(n-1)
仍然为等比数列
首项为 1/a1 ,公比为 1/q
S' = (1/a1) * [1 - (1/q)^n]/(1 - 1/q)
= (1/a1) * (q^n -1)/[(q-1)q^(n-1)]
(S/S')^n
= [(a1)^2 * q^(n-1)]^n
= a1^(2n) * q^[n(n-1)]
所以
T^2=(S/S')^n
命题成立