已知各项都是正数的数列an,其前n项和为Sn,Sn=(√Sn-1+√a1)^2

sn-(根号s(n-1)+根号a1)^2=0
(根号sn -根号s(n-1)-根号a1)(根号sn +根号(s(n-1)+根号a1)=0
由题意得
根号sn =根号s(n-1) +根号a1
=根号s(n-2)+2根号a1
=根号s(n-3)+3根号a1
=...
=根号s1 +(n-1)根号a1
=n*根号a1
sn=a1*n^2
s(n-1)=a1*(n-1)^2
an=a1*n^2-a1*(n-1)^2 =2a1*n -a1
bn=(2a1(n+1)-a1)/(2a1 n-a1) +(2a1 n-a1)/(2a1(n+1)-a1)
=(2a1 n+a1)/(2a1 n-a1)+(2a1 n-a1)/(2a1 n+a1)
=(2n+1)/(2n-1) +(2n-1)/(2n+1)
=1+2/(2n-1)+1-2/(2n+1)
=2+2/(2n-1)-2/(2n+1)
由上式可以 求出Tn (注意到2n-1 ,2n+1 相差2)