数列{an}满足an+1=1/(2-an),用a1,n(n>2)表示an=?
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数列{an}满足an+1=1/(2-an),用a1,n(n>2)表示an=?
数列{an}满足an+1=1/(2-an),
用a1,n(n>2)表示an=?
数列{an}满足an+1=1/(2-an),用a1,n(n>2)表示an=?
(an+1)-1=1/[2-an]-1=(an-1)/(2-an),
设bn=an-1,则b(n+1)=bn/(1-bn),
1/(bn+1)=1/bn-1,
1/(bn+1)-1/bn=-1,
故{1/bn}为首项是1/b1=1/(a-1),公差为-1的等差数列,
1/bn=1/(a-1)-(n-1),
bn=(a-1)/[n(1-a)+a],
所以an=bn+1=[(n-1)(1-a)]/[n(1-a)+a]
你应该把题重新输一遍的,这种情况,算题的时候没法算。到底是a下底的n+1还是a下底的n再加1呢?
1、演算几个:a2,a3,a4,。。。
2、an=[(n-1)-(n-2)a1]/[n-(n-1)a1](n>=2)
3、归纳法可证明。
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