(1)当n≥2时,cn=an+bn=+=an-1+bn-1+2, ∴cn=cn-1+2,即cn-cn-1="2" (n≥2) ∴数列{cn}为等差数列,首项c1=a1+b1=3,公差d=2. ∴cn=3+(n-1)×2=2n+1. (2)当n≥2时, ①-②得:an-bn=(an-1-bn-1) (n≥2), ∴数列{an-bn}为等比数列,首项为a1-b1=1,公比q=, ∴an-bn=()n-1. ③ 由(1)知:an+bn="2n+1, " ④ ③+④得2an="(2n+1)+" ()n-1 ∴an=+ ∴Sn=++…++ = =. |