bn=an^2-1分之1=1/[(an+1)(an-1)]=1/2(an-1)-1/2(an+1) 前n项之和为：1/2(an+1)-1/2(an-1)+1/2(an)-1/2(an-2)+1/2(an-1)-1/2(an-3)+.+1/2a3-1/2a1+1/2a2-1/2a0=1/2(an+1)+1/2an-1/2a0=1/[2*4^(n+1)]+1/(2*4n^2)-1/2*...