一道数列和不等式结合的题目数列｛an｝为等差数列,an为正整数,其前n项和为Sn,数列｛bn｝为等比数列.且a1=3,b1=1,数列｛ban｝（an是在b的下面）是公比为64的等比数列.b2×S2=64求证1/S1+2/S2+...+1/Sn

a(n)=3+(n-1)d,d为非负整数.
S(n)=3n+n(n-1)d/2.
b(n)=q^(n-1),
b[a(n)]=q^[a(n)-1]=q^[3+(n-1)d-1]=q^(3-1)*[q^d]^(n-1)=[q^2]*[q^d]^(n-1).
q^d=64.
64=b(2)S(2)=q*[6+d],
q=64/(6+d).
64=q^d=[64/(6+d)]^d=64^d/(6+d)^d,
(6+d)^d=64^(d-1)=2^[6(d-1)],
d=2.
S(n)=3n+n(n-1)d/2=3n+n(n-1)=n^2+2n=n(n+2)
1/S(n)=1/[n(n+2)]=(1/2)[1/n-1/(n+2)],
1/S(1)+1/S(2)+1/S(3)+...1/S(n-2)+1/S(n-1)+1/S(n)
=(1/2)[1/1-1/3 + 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+1/2 - 1/(n+1) - 1/(n+2)]
=(1/2)[3/2 - 1/(n+1) -1/(n+2)]
=3/4 -(1/2)[1/(n+1)+1/(n+2)]

(1) 由an 是等差数列知:a1=3,a2=3+d ；s2=a1+a2=6+d
bn是等比数列得：bn=q^(n-1),b2=q,又由
b2×S2=64 即q*（6+d）=64 记为一式；
b（an）是等比数列：b（a2）/b（a1）=q^d=64 记为二式
结合一二式解得：d=2,q=8
所以得：an=2n+1 ,bn=8^(n-1)
(2)由（1）知：sn=（a1+an）*n/2=n*(n+2)；
1/S1=1/(1*3) 1/2 *(1-1/3) ,1/S2=1/(2*4)=1/2*(1/2-1/4) ……
1/sn=1/（n*(n+2)）=1/2 *(1/n—1/n+2)
所以（2）
原式=1/2*(1-1/3 +1/2-1/4+1/3-1/5 +……+1/(n-1)-1/(n+1)+1/n -1/(n+2)
=1/2*(1+1/2-1/(n+1)-1/(n+2))
=1/2*(n/(n+1)+n+1/(n+2)-1/2)