在数列{an}中,an+Sn=n2+2n-1,n属于N* 令bn=an*(1/2)的n-1次方,证：b1+b2+b3+.

在数列{an}中,an+Sn=n2+2n-1,n属于N* 令bn=an*(1/2)的n-1次方,证：b1+b2+b3+.+bn

证明：∵an+Sn=n²+2n-1,∴a(n+1)+S(n+1)=(n+1)²+2(n+1)-1
则a(n+1)-an+S(n+1)-Sn=(n+1)²+2(n+1)-1-(n²+2n-1)
有2a(n+1)-an=2n+3 ； 整理：2[a(n+1)-2(n+1)+1]=an-2n+1
∵a1+S1=1²+2*1-1 ；∴2a1=2 ；则a1=1；a1-2*1+1=0
则2[a(n+1)-2(n+1)+1]=an-2n+1=0 ；故an=2n-1 n∈N* ；
则bn=an*(1/2)^(n-1)=(2n-1)*(1/2)^(n-1) n∈N* ；
又bn=(2n-1)*(1/2)^(n-1)=2n*(1/2)^(n-1)-(1/2)^(n-1)
则bn+(1/2)^(n-1)=n*(1/2)^(n-2)=(1/2)^(n-2)+(1/2)^(n-2)+•••+(1/2)^(n-2)-(1/2)^(n-1) 即n个(1/2)^(n-2)相加
则b1+b2+b3+•••+bn+(1/2)^(1-1)(1-1/2^n)/(1-1/2)=(1/2)^(1-2)+[(1/2)^(2-2)+(1/2)^(2-2)]+[(1/2)^(3-2)+(1/2)^(3-2)+(1/2)^(3-2)]+•••+[(1/2)^(n-2)+(1/2)^(n-2)+•••+(1/2)^(n-2)]=[(1/2)^(1-2)+(1/2)^(2-2)+(1/2)^(3-2)+•••+(1/2)^(n-2)]+[(1/2)^(2-2)+(1/2)^(3-2)+•••+(1/2)^(n-2)]+[(1/2)^(3-2)+1/2)^(4-2)+•••+(1/2)^(n-2)]+•••+[(1/2)^(n-3)+(1/2)^(n-2)]+(1/2)^(n-2)=2(1-1/2^n)/(1-1/2)+[1-1/2^(n-1)]/(1-1/2)+1/2[1-1/2^(n-2)]/(1-1/2)+•••+(1/2)^(n-3)(1-1/2²)/(1-1/2)+(1/2)^(n-2)(1-1/2)/(1-1/2)=2{1/2^(-1)(1-1/2^n)+1/2^0[(1-1/2^(n-1)]+1/2^1[1-1/2^(n-2]+•••+(1/2)^(n-2)(1-1/2)/}=2[1/2^(-1)+1/2^(0)+1/2^1+•••+(1/2)^(n-2)-n/2^(n-1)]=2[4(1-1/2^n)-n/2^(n-1)]=8-1/2^(n-3)-n/2^(n-2)
故b1+b2+b3+•••+bn+(1/2)^(1-1)(1-1/2^n)/(1-1/2)
=b1+b2+b3+•••+bn+2-1/2^(n-1)=8-1/2^(n-3)-n/2^(n-2)
则：b1+b2+b3+•••+bn=6+1/2^(n-1)-1/2^(n-3)-n/2^(n-2)=6-3/2^(n-3)-n/2^(n-2)