Question

设等比数列｛an}的前n项和为Sn,若S6/S3=3,则S9/S6=

Answer 1

若公比q=1，则S6/S3=(6a1)/(3a1)=2，与已知不符，因此q不等于1。
S6/S3=[a1(q^6 -1)/(q-1)]/[a1(q^3 -1)]/(q-1)
=(q^6 -1)/(q^3 -1)
=(q^3+1)(q^3-1)/(q^3-1)
=q^3+1=3
q^3=2

S9/S6=[a1(q^9-1)/(q-1)]/[a1(q^6-1)/(q-1)]
=(q^9 -1)/(q^6 -1)
=[(q^3)^3 -1]/[(q^3)^2 -1]
=(2^3 -1)/(2^2 -1)
=7/3

Answer 2

S6/S3
a1(1+q+q^2+q^3+q^4+q^5)/a1(1+q+q^2)=3
[(1+q+q^2)+q^3(1+q+q^2)]/(1+q+q^2)=3

1+q^3=3
q^3=2

同理S9/S3=(1+q+q^2+q^3+q^4+q^5+q^6+q^7+q^8)/(1+q+q^2)
=[(1+q+q^2)+q^3(1+q+q^2)+q^6(1+q+q^2)]/(1+q+q^2)
=1+q^3+q^6
=1+2+2^2=7
而S9/S6*S6/S3=S9/S3
S9/S6 * 3=7
S9/S6=7/3

Answer 3

解：设公比为q，首项为a1,
则S3=a1(1-q^3)/(1-q)
S6=a1(1-q^6)/(1-q)

∵ S6/S3=[a1(1-q^6)/(1-q)]/[a1(1-q^3)/(1-q)]
=(1-q^6)/(1-q^3)
=1+q^3=3
q^3=2

∴ S9/S6=[a1(1-q^9)/(1-q)]/[a1(1-q^6)/(1-q)]
=(1-q^9)/(1-q^6)
=(1-q^3)(1+q^3+q^6)/(1-q^3)(1+q^3)
=(1+q^3+q^6)/(1+q^3)
=(1+2+4)/(1+2)
=7/3
=

Answer 4

s6/s3=[a1(1-q^6)/(1-q)]/[a1(1-q^3)/(1-q)]
s6/s3=(1-q^6)/(1-q^3)

(1-q^6)/(1-q^3)=3

(1-q^3)(1+q^3)/(1-q^3)=3

1+q^3=3

q^3=2

S9/S6

=[a1(1-q^9)/(1-q)]/[a1(1-q^6)/(1-q)]
=(1-q^9)/(1-q^6)

=[1-(q^3)^3]/[1-(q^3)^2]
=[1-2^3]/[1-2^2]

=(-7)/(-3)
=7/3