Question

已知数列{a_n}的前n项和S_n满足a_n+2S_nS_n-1="0" (n≥2),a₁=,求a_n.

Answer 1

a_n=.

∵当n≥2时，a_n=S_n-S_n-1,
∴S_n-S_n-1+2S_nS_n-1=0，
即-=2，                                                       4分
∴数列是公差为2的等差数列.                                      6分
又S₁=a₁=，∴=2，
∴=2+（n-1）·2=2n，
∴S_n=.                                                               10分
∴当n≥2时，a_n=-2S_nS_n-1=-2··
=-，                                                            12分
∴a_n=.                                                14分