Decomposition factor: - 3ax-6ax-3ay,

Decomposition factor: - 3ax-6ax-3ay,

：-3ax-6axy-3ay =-3a(x +2xy+y) = -3a(x+y)

If S3 = 3, s9-s6 = 12, then S6=______ ．

The sum of the first n terms of {an} is Sn, S3, s6-s3 and s9-s6 are proportional sequences, that is, (s6-s3) 2 = S3 · (s9-s6), (s6-3) 2 = 3 × 12, so S6 = 9 or - 3 can be obtained

If S3 = 4, S6 = 12, then S9=

Let the first term be A1 and the common ratio be qs3 = a1 + A2 + a3 = A1 (1 + Q + Q & sup2;) s6-s3 = A4 + A5 + A6 = a1q & sup3; (1 + Q + Q & sup2;) s9-s6 = A7 + A8 + A9 = A7 + A8 + A9 = a1q ＾ 6 (1 + Q + Q & sup2;). We can see that (s6-s3) & sup2; = S3 (s9-s6) s9-s6 = (s6-s3) & sup2; (S3 = (12-4) & sup2; (4 = 16s9 =

In the equal ratio sequence {an}, if S6 / S3 = 3, find S9 / S6

S6/S3=[a1(q^6 -1)/(q-1)]/[a1(q^3 -1)/(q-1)]=q^3 +1=3
S9/S6=[a1(q^9 -1)/(q-1)]/[a1(q^6 -1)/(q-1)]
=(q^9-1)/(q^6 -1)
=[(q^3 -1)(q^6+q^3+1)]/[(q^3+1)(q^3 -1)]
=(q^6 +q^3 +1)/(q^3 +1)
=[(q^3+1)^2 -q^3]/(q^3+1)
=[(q^3+1)^2 -(q^3+1)+1]/(q^3+1)
=(9-3+1)/3
=7/3