急急急~~！！数学

注意到an^2=2(a(n-1))^2+1,
因此an^2+1=2[(a(n-1))^2+1]
因此an^2+1为首项为2，公比为2的等比数列
有桐兄：an^2+1=2^n,an=√(2^n-1)
因此,bn=2^n/[√(2^n-1)+√(2^(n+1)-1)]
设2^n=k，则bn=k/[√(k-1)+√(2k-1)]
=√(2k-1)-√(k-1)]
=√[(2^(n+1)-1]-√(2^n-1)
所以，Sn=b1+b2+...+bn
=（√3-1)+(√5-√3)+（√7-√5）+...+[√[(2^(n+1)-1]-√(2^n-1)]
=√[(2^(n+1)-1]-1
PS:一般非等比或等差的数列要注意变形化成等比等差数列来局戚袭处理，仔悄写出数列的通项公式一般对解题也会有所帮助。希望能帮助到你

解：
由已知有An²=2A(n-1)²+1,可设Cn=An²,则

Cn = 2C(n-1) + 1, C1 = 1

推出 Cn + 1 = 2[C(n-1) + 1] , C1 + 1 = 2

∴{Cn+1}为轮歼等比数列，易得 Cn + 1 = 2^n

∴ An = √Cn = √(2^n-1)

因此 Bn = 2^n / (An + A(n+1))

= 2^n (An-A(n+1)) / (An²-A(n+1)²)

= 2^n (An-A(n+1)) / (2^n-1-2^(n+1)+1)

= - (An-A(n+1))

故慧差有 Sn = ∑腊碧冲Bn

= -[(A1-A2) + (A2-A3) + (A3-A4) + …… + (An-A(n+1))]

= -[A1 - A(n+1)]

= A(n+1) - A1

= √[2^(n+1)-1] - 1