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