Question

设等比数列{an}的前n项和为Sn S4=1 S8=17求通项公式an

Answer 1

解：
设公比为q，若q=1，则S4=4a1 S8=8a1
S8/S4=8/4=2≠17，与已知不符，因此q≠1
S8/S4=[a1(q^8 -1)/(q-1)]/[a1(q^4-1)/(q-1)]=(q^4+1)(q^4-1)/(q^4-1)=q^4+1=17/1=17
q^4=16
q=2或q=-2
q=2时，S4=a1(q^4 -1)/(q-1)=15a1=1 a1=1/15
an=a1q^(n-1)=(1/15)×2^(n-1)
q=-2时，S4=a1(q^4-1)/(q-1)=-5a1=1 a1=-1/5
an=a1q^(n-1)=(-1/5)×(-2)^(n-1)=(-1)ⁿ×2^(n-1) /15

Answer 2

an= a1q^(n-1)
S4 = a1q^3 =1 (1)
S8= a1q^7 = 17 (2)
(2)/(1)
q^4= 17
q = (17)^(1/4) or -(17)^(1/4)
q= (17)^(1/4) a1= (17)^(-3/4)
q= -(17)^(1/4) a1= -(17)^(-3/4)

an = (17)^((n+2)/4) or (17)^(-(n+2)/4)