问题 填空题
已知对任意正整数n都有a1+a2+…+an=n3,则
1
a2-1
+
1
a3-1
+…+
1
a100-1
=______.
答案

∵a1+a2+a3+…+an=n3

∴a1=1,a1+a2=8,a1+a2+a3=27,a1+a2+a3+a4=64,a1+a2+a3+a4+a5=125,

∴a2=7,a3=19,a4=37,a5=61,an=3n(n-1)+1,

∴a100=3×100×99+1,

1
a2-1
+
1
a3-1
+…+
1
a100-1
=
1
6
+
1
18
+
1
36
+
1
60
+…+
1
3×100×99

=

1
3
1
2
+
1
6
+
1
12
+
1
20
+…+
1
100×99
),

=

1
3
(1-
1
2
+
1
2
-
1
3
+
1
3
-
1
4
+
1
4
-
1
5
+…+
1
99
-
1
100
),

=

1
3
(1-
1
100
),

=

33
100

故答案为:

33
100

单项选择题 A1/A2型题
计算题