∵an·a(n+1) + 2a(n+1) + 1 = 0
∴an·a(n+1) + a(n+1) + 1 = -a(n+1)
an·a(n+1) + an + a(n+1) + 1 = an - a(n+1)
(an + 1)·（a(n+1) + 1）= an - a(n+1)
即：
an - a(n+1)
———————————— = 1
(an + 1)·（a(n+1) + 1）
则：
1 1
—————— — ——————
a(n+1) + 1 an + 1
an + 1 - 【a(n+1) - 1】
= ————————————
(an + 1)·（a(n+1) + 1）
an - a(n+1)
= ————————————
(an + 1)·（a(n+1) + 1）
= 1
∵a1=1
∴1/(an + 1)=1/(1+1)=1/2
则数列{1/(an + 1)}为首项是1/2,公差是1的等差数列.