a²（n﹢1)-an·a(n＋1)-2a²n=0两边同时除以a²n[a(n+1)/an]²-a(n+1)/an-2=0解得：a(n+1)/an=-1或a(n+1)/an=2∵an>0∴a(n+1)/an=2那么{an}是等比数列,公比q=2∵a1=2∴an=2^n 设数列等差数列｛bn｝...
已知数列an的每一项都为正数,满足a1=2,且an﹢1∧2-an·an＋1-2an∧2=0等差数列bn的前n项和为Tn,
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