在等差数列an中，a1a3=0,a2a4=-2,则{an/2^n-1}前10项和为10

{a(n)}是等差数列，a(1)+a(3)=0，a(2)+a(4)=-2∴ 2d=[a(2)-a(1)]+[a(4)-a(3)]=[a(2)+a(4)]-[a(1)+a(3)]=-2∴ d=-1∴ a(1)+a(3)=a(1)+a(1)+2d=2a(1)-2∴ 2a(1)-2=0，a=1。∴ a(n)=2-n。b(n)=(1/2)ⁿa(n)-1s(n)=b(1)+b(2)+b(3)+……+b(n)s(n)+n=(1/2)a(1)+(1/2)²a(2)+(1/2)³a(3)+……+(1/2)ⁿa(n)①2[s(n)+n]=a(1)+(1/2)a(2)+(1/2)²a(3)+……+2(1/2)ⁿa(n)②②-①s(n)+n=a(1)-[(1/2)+(1/2)²+……+2(1/2)ⁿ]-(1/2)ⁿa(n)=a(1)-[1-2(1/2)ⁿ]-(1/2)ⁿa(n)=1-1+2(1/2)ⁿ-(1/2)ⁿ(2-n)=n(1/2)ⁿs(n)=n(1/2)ⁿ-n=n[(1/2)ⁿ-1]