题目内容
已知数列
满足递推关系式
,又
,则使得
为等差数列的实数
。
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823214550053480.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823214550084922.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823214550100409.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823214550131838.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823214550146346.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823214550162363.png)
解:设bn=(an+λ)/ 3n,根据题意得bn为等差数列即2bn=bn-1+bn+1,而数列{an}满足递推式an=3an-1+3n-1(n≥2),
可取n=2,3,4得到(3a1+32-1+λ)/ 32 +(3a3+34-1+λ)/ 34 =2(3a2+33-1+λ) /33,
而a2=3a1+32-1,a3=3a2+33-1=3(3a1+32-1)=9a1+33-3,代入化简得λ="-1" /2 .
故答案为:-1/ 2
可取n=2,3,4得到(3a1+32-1+λ)/ 32 +(3a3+34-1+λ)/ 34 =2(3a2+33-1+λ) /33,
而a2=3a1+32-1,a3=3a2+33-1=3(3a1+32-1)=9a1+33-3,代入化简得λ="-1" /2 .
故答案为:-1/ 2
![](http://thumb2018.1010pic.com/images/loading.gif)
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