設(shè)an的等差為b,bn的等比為d b2s2=64
有d(6+b)=64
b（an）是公比為64的等比數(shù)列 b(a2)/b(a1)=64
有 b(3+b)/b3=64 即 b^d=64
解得 b=2 d=8 所以 an=3+2(n-1) bn=8^(n-1)
Sn=n(n+2)
所以1/Sn=[(1/n)-1/(2+n)]/2
所以S=1/S1+1/S2+...+1/Sn
=(1/2)[(1-1/3)+(1/2-1/4)+(1/3-1/5)+……+(1/n)-1/(2+n)]
=（1/2）[1+1/2-1/(n+1)-1/(n+2)]
=3/4-[(2n+3)/2(n+1)(n+2)]
因為n為大于0的整數(shù)所以[(2n+3)/2(n+1)(n+2)] >0
所以S=1/S1+1/S2+...+1/Sn