数列{2n-1/2*n},求其前n项的和Sn

因为an=(2n-1)/2^n=n/2^(n-1)-1/2^n设数列{n/2^(n-1)}前n项和为Tn,数列{1/2^n}前n项和为Pn,则Sn=Tn-PnTn=1+2/2+3/2²+4/2³+.+n/2^(n-1)(1/2)Tn=1/2+2/2²+3/2³+.+(n-1)/2^(n-1)+n/2^n上两式错项相...