已知等差数列{an}满足:a3+a4=16,a4+a5=20,{an}的前n项和为Sn

⑴∵等差数列{an}满足a3+a4=16,a4+a5=20,
∴2a1+5q=16
2a1+7q=20
∴a1=3,q=2
∴Sn=na1+n(n-1)d/2=n²+2n
⑵1/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/(n-2)-1/n+1/(n-1)-1/(n+1)+1/n-1/(n+2)]
=1/2[1+1/2-1/(n+1)-1/(n+2)]
=3/4-1/[2(n+1)(n+2)]
【第二小问考察了裂项求和法,即1/n(n+2)=1/2[1/n-1/(n+2)],再消去中间部分.】
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