跪求此简算题的解!1/2+3/22+4/23+……11/2十次方=?

设S=1/2+3/2^2+4/2^3+……11/2^10,则
(1/2)S=1/2^2+3/2^3+……10/2^10+11/2^11,两式相减:
S-(1/2)S=(1/2)+(2/2^2)+1/2^3+1/2^4……1/2^10-11/2^11
(1/2)S=(1/2)+(1/2)+[1/2^3+1/2^4……1/2^10]-11/2^11
(1/2)S=1+[1/2^3+1/2^4……1/2^10]-11/2^11
(1/2)S=1+1/2^3[1-(1/2)^8/(1-1/2)]-11/2^11
(1/2)S=1+(1/2)^2*[1-(1/2)^8]-11/2^11
(1/2)S=1+(1/4)*(255/256)-11/2^11
(1/2)S=1+(255/1024)-11/2048
(1/2)S=1+(510/2048)-11/2048
(1/2)S=1+499/2048
S=2+499/1024
(2又499/1024)