已知数列 {an} 的前n项和为Sn=1/2n^2+1/2n n∈N*

Sn=1/2n^2+1/2na1=S1=1/2+1/2=1n>=2时an=Sn-S(n-1)=(1/2n^2+1/2n)-(1/2(n-1)^2+1/2(n-1))=n∵a1=1满足an=n,n>=2∴an=n,n∈N*bn=1/[(n+2)*an]=1/[(n+2)n]=1/2*[1/n-1/(n+2)]∴Tn=b1+b2+.+bn=1/2*(1-1/3+1/2-1/4+1/3-1...