Question

已知等比数列{an}的前n项和为Sn,且S4/S8=1/3,那么S8/S16的值为?

Answer 1

解：
设公比为q
若q=1，则S4/S8=(4a1)/(8a1)=1/2≠1/3，与已知矛盾，因此q≠1
S4/S8=1/3
[a1(q^4 -1)/(q-1)]/[a1(q^8 -1)/(q-1)]=1/3
(q^4 -1)/[(q^4 +1)(q^4 -1)]=1/3
1/(q^4 +1)=1/3
q^4 +1=3
q^4=2

S8/S16=[a1(q^8 -1)/(q-1)]/[a1(q^16 -1)/(q-1)]
=(q^8 -1)/[(q^8 +1)(q^8 -1)]
=1/(q^8 +1)
=1/[(q^4)²+1]
=1/(2²+1)
=1/5

Answer 2

S4/S8=1/3
[a1(1-q^4)/(1-q)]/[a1(1-q^8)/(1-q)]=1/3
(1-q^4)/(1-q^8)=1/3
(1-q^4)/[(1-q^4)(1+q^4)]=1/3
1/(1+q^4)=1/3
q^4+1=3
q^4=2
S8/S16
=[a1(1-q^8)/(1-q)]/[a1(1-q^16)/(1-q)]
=(1-q^8)/(1-q^16)
=(1-q^8)/[(1-q^8)(1+q^8)]
=1/(1+q^8)
=1/[1+(q^4)^2]
=1/(1+2^2)
=1/5

追问

抱歉，都回答的一样好，不太好抉择呢~

追答

没关系