an^2=2a(n-1)^2+1an^2+1=2(a(n-1)^2+1)考虑数列an^2+1,不难滑竖得到an^2+1=2^n,信侍大即an=√(2^n-1)将an代入bn=2^n/(an+a(n+1)),分子分母同乘√(2^(n+1)-1)-√(2^n-1),得到bn=√(2^(n+1)-1)-√(2^n-1),求和就很简单了,∑谈散bn=√(2^(n+1)-1)-1.