已知数列{an}满足a1=0,对任意k∈N*,有a2k-1 a2k a2k+1成公差为k的等差数列,数列bn=（2n+1

a(2k)-a(2k-1)=k (1)
a(2k+1)-a(2k)=k (2)
(1)+(2)
a(2k+1)-a(2k-1)=2k
a(2n+1)-a(2n-1)=2n
a(2n-1)-a(2n-3)=2(n-1)
…………
a3-a1=2
累加
a(2n+1)-a1=2(1+2+...+n)=2n(n+1)/2=n(n+1)
a(2n+1)=a1+n(n+1)=0+n(n+1)=n(n+1)
bn=(2n+1)²/a(2n+1)=(2n+1)²/[n(n+1)]=[(2n+1)/n][(2n+1)/(n+1)]
=(1/n +2)[(2n+2-1)/(n+1)]
=(1/n +2)[2 -1/(n+1)]
=2/n -1/[n(n+1)]+4 -2/(n+1)
=2/n -1/n +1/(n+1) -2/(n+1) +4
=1/n -1/(n+1) +4
Sn=b1+b2+...+bn=[1/1-1/2+1/2-1/3+...+1/n-1/(n+1)]+4n
=[1-1/(n+1)]+4n
=n/(n+1) +4n