Question

已知递增等比数列{an}的第3项，第5项，第7项的积为512，且这三项分别减去1，3，9后成等差数列

求{an}得首项和公比
设Sn=a1^2+a2^2+........+an^2,求Sn

Answer 1

设首项为a1，公比为q
a1*q^2*a1*q^4*a1*q^6=(a1*q^4)^3=512
a1*q^4=8 (1)

2*(8-3)=(a1*q^2-1)+(a1*q^6-9)
a1*q^2(1+q^4)=20 (2)

(1)/(2)得 q^2/(1+q^4)=8/20=2/5
5q^2=2+2q^4
2q^4-5q^2+2=0
(2q^2-1)(q^2-2)=0
q^2=1/2或q^2=2
代入（1）得
｛a1=16 或｛a1=16 或 ｛a1=4 或｛a1=4
｛q=-√2/2 ｛q=√2/2 ｛q=-√2 ｛q=√2

Sn=a1^2+a2^2+...+an^2
=a1^2(1+q^2+q^4+...+q^(2n))
=a1^2*(1-q^(2n+2))/(1-q^2)
=256*[1-(1/2)^(n+1)]/(1/2)=512*[1-(1/2)^(n+1)]
或
Sn=4*[1-2^(n+1)]/(1-2)=4*[2^(n+1)-1]