已知a1=1,b1=−1,an+1=anbn+1,bn+1=bn1−4a2n,求数列{an}和{bn}的通项公式.
分析与解 若消bn,则有an+1an=anan−11−4a2n,
即an+1a2n=1an−1(1−4a2n),
接下来无从下手,转而选择消去an.
根据已知,有4a2n=1−bnbn+1,
而4a2n+1=4a2nb2n+1,
从而1−bn+1bn+2=(bn+1−bn)⋅bn+1,
从而bn+1+1bn+2=bn+1bn+1,
因此数列{bn+1bn+1}为常数列,有bn+1bn+1=2,
于是1bn+1−1=1bn−1−1,
因此可以求得1bn−1=−n+12,
化简得bn=2n−32n−1.
进而an=bn⋅bn−1⋯b2⋅a1=12n−1.
综上,数列{an}和{bn}的通项公式为an=12n−1,bn=2n−32n−1.