a(n+1)=(nan)/[(n+1）*（nan+1)]
兩邊乘以n+1,在取倒數(shù)
1/[a(n+1)*(n+1)] = 1 + 1/(nan)
所以b(n+1) = 1 + bn ,即bn是等差數(shù)列（lz你應(yīng)該是題目輸錯了 bn=1/nan ）
(2)、由于bn是等差數(shù)列 bn = b1 + 1 * (n-1)
= 2 + n-1
= n+1
bn = 1/(nan) => an = 1/[n*(n+1)]
= 1/n - 1/(n+1)
Sn = 1 - 1/2 + 1/2 - 1/3 + …… + 1/n -1/(n+1)
= 1- 1/(n+1)
= n/(n+1)