Question

第四题求详解

Answer 1

选A吧
∵a8=2a4
∴a1 + 7d=2(a1 + 3d)
a1 + 7d=2a1 + 6d
∴a1=d
∵S5=5a1 + 10d=5a1 + 10a1
=15a1=15
∴a1=1=d，则an=n
∴Sn=n(1+n)/2
则1/Sn=2/n(n+1)

追答

Answer 2

S[5]=15 推出 a[3]=3;
a[5]=2a[4], 即 3+5d=2(3+d), 得到 d=1；
所以 a[n]=n;
S[n]=n(n+1)/2;
1/S[n]=2/[n(n+1)]=2/n-2/(n+1);
裂项求和等于 2n/(n+1)

Answer 3

S5=5a3=15
a1+2d=3 ①
a8=2a4
a1+7d=2(a1+3d)
a1-d=0 ②
联立①、②，解得a1=1，d=1
1/Sn=1/[na1+n(n-1)d/2]
=1/[n·1+n(n-1)·1/2]
=2/[n(n+1)]
=2[1/n -1/(n+1)]
1/S1+1/S2+...+1/Sn
=2[1/1 -1/2 +1/2 -1/3+...+1/n -1/(n+1)]
=2[1- 1/(n+1)]
=2n/(n+1)
选A