姚智灵
幼苗
共回答了16个问题采纳率:75% 举报
设第k行的第一个数为ak,
则a
1 =1,
a
2 =4=2a
1 +2,
a
3 =12=2a
2 +2
2 ,
a
4 =32=2a
3 +2
3 ,
…
由以上归纳,得a
k =2a
k-1 +2k-1(k≥2,且k∈N*),
∴a
k ![](https://img.yulucn.com/upload/d/f3/df3636bc1ce24ae013c9625341c2cf40_thumb.jpg)
2
k ="a"
k-1 ![](https://img.yulucn.com/upload/d/f3/df3636bc1ce24ae013c9625341c2cf40_thumb.jpg)
2
k-1 +
![](https://img.yulucn.com/upload/d/53/d53e20b1aa26546cd7ed0af0bf94e36d_thumb.jpg)
,
∴数列{
![](https://img.yulucn.com/upload/5/43/543fad1e961c6f4dfc5da683fbdaeff0_thumb.jpg)
}是以
![](https://img.yulucn.com/upload/d/53/d53e20b1aa26546cd7ed0af0bf94e36d_thumb.jpg)
为首项,以
![](https://img.yulucn.com/upload/d/53/d53e20b1aa26546cd7ed0af0bf94e36d_thumb.jpg)
为公差的等差数列,
∴
![](https://img.yulucn.com/upload/5/43/543fad1e961c6f4dfc5da683fbdaeff0_thumb.jpg)
=1
![](https://img.yulucn.com/upload/d/53/d53e20b1aa26546cd7ed0af0bf94e36d_thumb.jpg)
+(n-1)×
![](https://img.yulucn.com/upload/d/53/d53e20b1aa26546cd7ed0af0bf94e36d_thumb.jpg)
,
∴an=n•2
n-1 (n∈N*).
由数阵的排布规律可知,每行的数(倒数两行另行考虑)都成等差数列,
且公差依次为:2,22,…,2k,…
第n行的首项为an=n•2
n-1 (n∈N*),公差为2
n ,
∴第32行的首项为a
32 =32•2
31 =2
36 ,公差为2
32 ,
∴第32行的第17个数是2
36 +16×232=2
37 .
故答案为:2
37 .
1年前
7