⑴∵等差数列{an}满足a3+a4=16,a4+a5=20,∴2a1+5q=162a1+7q=20∴a1=3,q=2∴Sn=na1+n(n-1)d/2=n²+2n⑵1/Sn=1/(n²+2n)=1/n(n+2)=1/2[1/n-1/(n+2)]∴Tn=1/S1+1/S2+...+1/Sn=1/2[1-1/3+1/2-1/4+1/3-1/5+...+1/...非常感谢