数列｛an｝,a1=8,a4=2,且满足an+2=2an+1-an,设Sn=|a1|+|a2|+...+|an|,求Sn

∵an+2＝2an+1－an ∴an+2－an+1＝an+1－an＝…＝a2－a1
∵a3＝2a2－a1 ∴a4＝2a3－a2＝3a2－2a1 ∴3a2＝2＋2×8 ∴a2＝6
∴an+2－an+1＝an+1－an＝…＝a2－a1＝﹣2
∴(an－an-1)＋(an-1＋an-2)＋…＋(a2－a1)＝﹣2(n－1)
∴an－a1＝﹣2(n－1)
∴an＝10－2n
∴n＜6 时，an≥0 ∴Sn＝|a1|＋|a2|＋…＋|an|＝a1＋a2＋…＋an＝(8＋10－n)n/2＝n(9－n)
n≥6时，an＜0 ∴Sn＝|a1|＋|a2|＋…＋|an|＝a1＋a2＋…＋a5－(a6＋a7＋a8＋…＋an)
＝﹣(a1＋a2＋…＋an)＋2×(a1＋a2＋…＋a5)
＝﹣n(9－n)＋2×20
＝n²－9n＋40