(I)因为(nÎN),易根据等差数列的定义判断出{yn}为等差数列. (II)解本小题的关键是先根据xn+1-xn=2为常数,可确定的奇数项和偶数项分别成等差数列,从而求出. (III) 要使AnBnAn+1为直角三形,则 |AnAn+1|=2=2()Þxn+1-xn=2(), 当n为奇数时,xn+1-xn=2(1-a);当n为偶数时,xn+1-xn=2a.然后分别研究即可. (1)(nÎN),yn+1-yn=,∴{yn}为等差数列 (4¢) (2)xn+1-xn=2为常数 (6¢) ∴x1,x3,x5,…,x2n-1及x2,x4,x6,,…,x2n都是公差为2的等差数列, ∴x2n-1=x1+2(n-1)=2n-2+a,x2n=x2+2(n-1)=2-a+2n-2=2n-a, ∴xn= (3)要使AnBnAn+1为直角三形,则 |AnAn+1|=2=2()Þxn+1-xn=2() 当n为奇数时,xn+1=n+1-a,xn=n+a-1,∴xn+1-xn=2(1-a). Þ2(1-a)=2() Þa=(n为奇数,0<a<1) (*) 取n=1,得a=,取n=3,得a=,若n≥5,则(*)无解; (14¢) 当偶数时,xn+1=n+a,xn=n-a,∴xn+1-xn=2a. ∴2a=2()Þa=(n为偶数,0<a<1) (*¢),取n=2,得a=, 若n≥4,则(*¢)无解. 综上可知,存在直角三形,此时a的值为、、. (18¢) |