高二数学-已知数列『an』中a1=2,a(n+1)=an+2n...若an+3n-2=2/bn,求数列bn的前n项和Sn.

a(n+1)=an +2n
a(n+1)-an=2n
an-a(n-1)=2(n-1)
a(n-1)-a(n-2)=2(n-2)
…………
a2-a1=2
累加
an-a1=2[1+2+...+(n-1)]=2n(n-1)/2=n²-n
an=a1+n²-n=2+n²-n=n²-n+2
2/bn=an+3n-2=n²-n+2+3n-2=n²+2n
bn=2/(n²+2n)=2/[n(n+2)]=1/n -1/(n+2)
Sn=b1+b2+...+bn
=1/1-1/3+1/2-1/4+...+1/n-1/(n+2)
=(1/1+1/2+...+1/n)-[1/3+1/4+...+1/(n+2)]
=1+1/2 -1/(n+1)-1/(n+2)
=3/2 -1/(n+1) -1/(n+2)