即 an 22 1(n N*). (II)证法一: 414
b 1b2 1
...4bn 1 (an 1)bn.
4(b1 b2 ... bn) n 2nbn.
2[(b1 b2 ... bn) n] nbn, ①
2[(b1 b2 ... bn bn 1) (n 1)] (n 1)bn 1. ② ②-①,得2(bn 1 1) (n 1)bn 1 nbn, 即(n 1)bn 1 nbn 2 0,
nbn 2 (n 1)bn 1 2 0.
④-③,得 nbn 2 2nbn 1 nbn 0, 即 bn 2 2bn 1 bn 0,
b*n 2 bn 1 bn 1 bn(n N),
bn 是等差数列。
③
④