设正项等比数列{an}的首项a1=0.5,前n项和为sn,且2^10S30-(2^10+1）S20+S20=0

S30=a1*(q^30-1)/(q-1)=a1(q^10-1)(q^20+q^10+1)/(q-1)
S20=a1*(q^20-1)/(q-1)=a1(q^10-1)(q^10+1)/(q-1)
S10=a1*(q^10-1)/(q-1)
2^10*a1(q^10-1)(q^20+q^10+1)/(q-1)-(2^10+1)*a1(q^10-1)(q^10+1)/(q-1)+a1*(q^10-1)/(q-1)=0
2^10*(q^20+q^10+1)-(2^10+1)(q^10+1)+1=0
2^10*q^20+2^10*q^10+2^10-2^10*q^10-2^10-q^10-1+1=0
2^10*q^20-q^10=0
所以q^10=1/2^10
各项均为正值
q>0
q=1/2
an=1/2*(1/2)^(n-1)=(1/2)^n
Sn=1/2*[1-(1/2)^n]/(1-1/2)=1-(1/2)^n
nSn=n-n*(1/2)^n
Tn=[1-1*(1/2)]+[2-2*(1/2)^2]+……+[n-n*(1/2)^n]
=1+……+n-[1*(1/2)+2*(1/2)^2+……+n*(1/2)^n]
令x=1*(1/2)+2*(1/2)^2+……+n*(1/2)^n
2x=1+2*(1/2)+……+n*(1/2)^(n-1)
x=2x-x=1+1*(1/2)+1*(1/2)^2+……+1*(1/2)^(n-1)-n*(1/2)^n
=1*[1-(1/2)^n]/(1-1/2)]-n*(1/2)^n
=2-2*(1/2)^n-n*(1/2)^n
=2-(n+2)(1/2)^n
所以Tn=1+……+n-x
=n(n+1)/2-2+(n+2)(1/2)^n