已知a1=2,点（an,a(n+1)）在函数f(x)=x^2+2x的图像上,其中n=1,2,...,
则f(an)=(an)^2+2(an)=a(n+1),
显然an>0,a(n+1)=an(an+2).
bn=(1/an)+[1/a(n+2)]
数列{bn}的前n项和Sn
Sn=b1+b2+b3+.+bn=
=（1/a1+1/a3)+(1/a2+1/a4)+(1/a3+1/a5)+...+[1/an+1/a(n+2)]=
={1/a1+1/[(a2)*(a2+2)]}+{1/[(a1)*(a1+2)]+1/[(a3)*(a3+2)]}+.+{1/[(a(n-1))^2+2(a(n-1))]+1/[(a(n+1))^2+2(a(n+1))]},
因为 1/[(a1)*(a1+2)]=（1/2）[1/(a1)-1/(a3)],
1/[(a2)*(a2+2)]=（1/2）[1/(a2)-1/(a4)],
1/[(a3)*(a3+2)]=（1/2）[1/(a3)-1/(a5)],
.,
1/[(a(n-1))*(a(n-1)+2)]=（1/2）[1/(a(n-1)-1/(a(n+1))],
1/[(an))*(an+2)]=（1/2）[1/(an)-1/(a(n+2))],
1/[(a(n+1))*(a(n+1)+2)]=（1/2）[1/(a(n+1)-1/(a(n+3))],
所以,
Sn=1/a1+（1/2）[1/(a2)-1/(a4)]+（1/2）[1/(a1)-1/(a3)]+
+（1/2）[1/(a3)-1/(a5)]+.+（1/2）[1/(a(n-1)-1/(a(n+1))]+
+（1/2）[1/(a(n+1)-1/(a(n+1))]
=（3/2)(1/a1)+(1/2)(1/a2)-(1/2)[1/(a(n+3))]
=（3/2)(1/2)+(1/2)(1/8)-(1/2)[1/(a(n+3))]
=13/16-(1/2)[1/(a(n+3))]
