设等比数列an的各项均为正值,首项a1=1/2,前n项和为Sn,且2^10S30-(2^10+1)S20+S10=0(1

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  • 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