(1)Sn=2n^2+nS(n-1)=2(n-1)^2+n-1 =2n^2-4n+2+n-1 =2n^2-3n+1Sn-S(n-1)=an=2n^2+n-2n^2+3n-1=4n-1 an=4log2 (bn)+3=4n-14log2(bn)=4n-4log2(bn)=n-1bn=2^(n-1) (2)an*bn=(4n-1)2^(n-1)=4n*2^(n-1)-2^(n-1)=n2^(n-1+2)-2^(n-1)=n2^(n+1)-2^(n-1)Tn=1*2^2-2^0+2*2^3-2^1+................+n2^(n+1)-2^(n-1)=1*2^2+2*2^3+...........+n2^(n+1)-(2^0+2^1+..........+2^(n-1))∵2^0+2^1+........+2^(n-1)是首项k1=1，公比q=2的等比数列2^0+2^1+........+2^(n-1)=1*(2^n-1)/(2-1)=2^n-1 设Mn=1*2^2+2*2^3+...........+n2^(n+1)2Mn=2*2^2+2*2^4+.............+n2^(n+2)两式相减得2Mn-Mn=Mn=1*2^3+2*2^4+.............+n2^(n+2)-1*2^2-2*2^3-...........-n2^(n+1)=-2^2-2^3-............-2^(n+1)+n2^(n+2)∵数列-2^2-2^3-............-2^(n+1)是首项=-4，公比q=2的等比数列-2^2-2^3-............-2^(n+1)=-4(2^n-1)(2-1)=-4(2^n-1)所以Mn=n2^(n+2)-4(2^n-1) =n2^(n+2)-2^(n+2)+4 =(n-1)2^(n+2)+4
