(本小题14分) (Ⅰ)由f(x)=a•2x+b的图象经过A(1,1),B(2,3)两点⇒⇒, ∴f(x)=2x-1, 又C(n,Sn)在f(x)的图象上⇒Sn=2n-1, 当n=1时,a1=S1=1; 当n≥2时,an=Sn-Sn-1=2n-1, ∴an=2n-1. (Ⅱ)由(Ⅰ)知, cn=3n•2n-n⇒Tn=3(1•21+2•22+…+n•2n)-(1+2+3+…+n), 令Pn=1•21+2•22+…+n•2n, 由错位相减法可求得Pn=(n-1)2n+1+2, 又1+2+3+…+n=, 故Tn=3Pn-=3(n-1)2n+1+6-. (Ⅲ)由Tn-=3(n-1)2n+1+6--=6(n-1)[2n-(2n+1)] 当n=1时,6(n-1)[2n-(2n+1)]=0,Tn= 当n=2时,6(n-1)[2n-(2n+1)]=-6,Tn< 当n=3时,6(n-1)[2n-(2n+1)]=12,Tn> 下证n≥3时,Tn>, 即证n≥3时,2n>2n+1, ∵n≥3时,2n=(1+1)n=+++…++≥++…++=2n+2>2n+1成立, ∴n≥3时,Tn>成立, 综上所述:n=1时,Tn=; n=2时,Tn<; n≥3时,Tn>. |