因为{Sn +1}是公比为2的等比数列,设首项为a所以 Sn +1=a2^(n-1)Sn =a2^(n-1) -1n≥2时,有an=Sn-Sn-1=(a2^(n-1) -1)-[a2^(n-2)-1]=a2^(n-1)-a2^(n-2)=a2^(n-2)所以an=a2^(n-2)又a1=S1=a2^(1-1) -1=a-1又an=a2^(n-2)所...数列{nan}的前n项和Tn怎么求呢？Tn=1*2^0+2*2^1+3*2^2+......+n*2^(n-1)2Tn=1*2^1+2*2^2+3*2^2+....+n*2^n两式错项相减Tn-2Tn=1+2^1+2^2+2^3+....+2^(n-1)-n*2^n-Tn=(2^n-1)/(2-1)-n*2^nTn=-2^n+1+n*2^n=(n-1)2^n+1