1 |
2 |
1 |
3 |
2 |
3 |
1 |
4 |
2 |
4 |
3 |
4 |
1 |
5 |
2 |
5 |
3 |
5 |
4 |
5 |
1 |
6 |
发现他们的个数是1,2,3,4,5…
构建新数列bn,则bn=
1 |
2 |
利用等差数列的和知道T5=
15 |
2 |
21 |
2 |
所以ak定在
1 |
7 |
2 |
7 |
3 |
7 |
6 |
7 |
又因为Sk<10,Sk+1≥10,而T5+
1 |
7 |
2 |
7 |
3 |
7 |
4 |
7 |
5 |
7 |
9 |
14 |
T5+
1 |
7 |
2 |
7 |
3 |
7 |
4 |
7 |
5 |
7 |
6 |
7 |
1 |
2 |
所以ak=
5 |
7 |
故答案为:
5 |
7 |
1 |
2 |
1 |
3 |
2 |
3 |
1 |
4 |
2 |
4 |
3 |
4 |
1 |
5 |
2 |
5 |
3 |
5 |
4 |
5 |
1 |
6 |
1 |
2 |
1 |
3 |
2 |
3 |
1 |
4 |
2 |
4 |
3 |
4 |
1 |
5 |
2 |
5 |
3 |
5 |
4 |
5 |
1 |
6 |
1 |
2 |
15 |
2 |
21 |
2 |
1 |
7 |
2 |
7 |
3 |
7 |
6 |
7 |
1 |
7 |
2 |
7 |
3 |
7 |
4 |
7 |
5 |
7 |
9 |
14 |
1 |
7 |
2 |
7 |
3 |
7 |
4 |
7 |
5 |
7 |
6 |
7 |
1 |
2 |
5 |
7 |
5 |
7 |