（1）当bn = b = 1/2时,b(n+1) = b,所以对任意n,bn = b = 1/2
（2）b(n+1) = 2bn^2 - bn + 1/2
所以b(n+1) - 1/2 = 2bn^2 - bn
1/[ b(n+1) - 1/2 ] = 1/[ 2bn (bn - 1/2) ]
1/[ b(n+1) - 1/2 ] = 1/[ bn - 1/2 ] - 1/bn
所以 1/bn = 1/[ bn - 1/2 ] - 1/[ b(n+1) - 1/2 ]
1/b1+1/b2+1/b3+.+1/bn = 1/[ b1 - 1/2] - 1/[ b(n+1) - 1/2 ] < 1/[ b -1/2] = 2/(2b-1)第一小题是b1=b啊！不是bn=b归纳法请问您如何证明是等差的常数列按您这样的话大家都是常数列了！