sn为数列{an}的前n项和,已知an>0,an^2+2an=4sn+3

sn为数列{an}的前n项和,已知an>0,an^2+2an=4sn+3

(1)根据an^2+2an=4Sn+3有：a(n+1)^2+2a(n+1)=4S(n+1)+3于是an^2+2an = a(n+1)^2+2a(n+1)-4a(n+1)=a(n+1)^2-2a(n+1)(an+1)^2 = [a(n+1)-1]^2化简得到a(n+1) = -ana(n+1) = an +2因为an>0,所以只有a（n+1) = an+2 满足要求，也就是他是等差数列又因为n=1时，a1^2 +2a1 = 4a1+3,a1 = 1an = 1 + 2(n-1)=2n-1(2)bn = 1/(2n-1)(2n+1) = 0.5 *[1/(2n-1) -1/(2n+1)]Sbn = b1 + b2 +....+bn= 0.5(1/1-1/3) + 0.5(1/3-1/5) +....+0.5[1/(2n-1) -1/(2n+1)]=0.5-0.5/(2n+1)