4Sn=(an+1)^2
4S(n-1)=(a(n-1)+1)^2
[注：S和a后面的括號(hào)中為下腳標(biāo)]
4an＝4Sn-4S(n-1)= (an+1)^2 -(a(n-1)+1)^2
所以：(an-1)^2＝(a(n-1)+1)^2
an-1＝±(a(n-1)+1)
由于為{an}為正項(xiàng)數(shù)列,所以an-1＝a(n-1)+1
an-a(n-1)＝2
為等差數(shù)列,公差為2
an=1+2(n-1)＝2n-1
bn=(2n-1)*(1/2)^n
Tn=1*(1/2)^1+3*(1/2)^2+5*(1/2)^3+……+(2n-3)*(1/2)^(n-1)+(2n-1)*(1/2)^n
Tn/2＝1*(1/2)^2+3*(1/2)^3+5*(1/2)^4+……+(2n-3)*(1/2)^n+(2n-1)*(1/2)^(n+1)
兩式相減：
Tn/2
＝1/2+2*[(1/2)^2+(1/2)^3+(1/2)^4+……+(1/2)^n]-(2n-1)*(1/2)^(n+1)
＝3/2-(1/2)^(n-1)-(2n-1)*(1/2)^(n+1)
＝3/2-(2n+3)*(1/2)^(n+1)
Tn=3-(2n+3)*(1/2)^n