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递推数 列

递推数 列. 1 、 a n +1 = a n +  ( n ) 型:. 方法:由 a n +1 = a n +  ( n ) 得: a n +1  a n = ( n ) ,令 n =1 , 2 , 3 ,. n  1 相加即得,这种方法称为累加法。. 例 1 、设数列 { a n } : a 1 =  2 ,以后各项由公式 a n = a n  1 + n 确定,求该数列的通项公式。. 练习:在数列 { a n } 中, 求 { a } 的通项公式。. 2 、 a n +1 = a n  ( n ) 型:.

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递推数 列

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  1. 递推数列

  2. 1、an+1=an+(n)型: 方法:由an+1=an+(n)得:an+1an=(n),令n=1,2,3, n1相加即得,这种方法称为累加法。 例1、设数列{an}:a1= 2,以后各项由公式an=an1+n确定,求该数列的通项公式。 • 练习:在数列{an}中, • 求{a}的通项公式。

  3. 2、an+1=an(n)型: =(n)令n=1,2,3, 方法:由an+1=a(n)得: n1相乘即得,这种方法称为累乘法。 例2、设数列{an}有:a1=2,此数列以后各项由 公式an= an1给出,求该数列的通项公式。

  4. 练习:在数列{an}中, (1)求{an}的通项公式; (2)令bn= ,求此数列的前n项的和。

  5. 3、an+1=pan+(n)型:(p是常数且p1) 方法:利用待定系数法构造等比数列((n)是常数,简捷!),或同时除以pn而化为题型1或等差数列、等比数列求得通项公式;也可除以(n)(f(n)不是常数),转化为上述题型处理。 例3、设数列{an}中,a1=2,且an+1=2an+3,求该数列的通项公式。 思考、设数列{an}中,a1=2,且an+1=2an+3n,求该数列的通项公式。 • 练习:1、已知数列 满足: a1=1,an+1=2an+32n1,求{an}的通项公式。

  6. 求数列通项公式常用技巧: 1、取倒数: an+1an+pan+1pan=0型都可这样处理,注意特点

  7. 3、取对数:

  8. 4、解方程(组): 例4、设{an}是首项为1的正项数列,且 (n+1)an+12nan2+an+1an=0,(n=1,2,3,…), 求它的通项公式。

  9. 5、数列与函数的关系:

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