已知对于任意正整数n,都有a1+a2+…+an=n3,则1/a2−1+1/a3−1+…+1/a100−1=_.

问题描述:

已知对于任意正整数n,都有a1+a2+…+an=n3,则

1
a2−1
+
1
a3−1
+…+
1
a100−1
=______.

∵当n≥2时,有a1+a2+…+an-1+an=n3,a1+a2+…+an-1=(n-1)3,两式相减,得an=3n2-3n+1,

1
an−1
=
1
3n(n−1)
=
1
3
1
n−1
-
1
n
),
1
a2−1
+
1
a3−1
+…+
1
a100−1

=
1
3
(1-
1
2
)+
1
3
1
2
-
1
3
)+…+
1
3
1
99
-
1
100
),
=
1
3
(1-
1
100
),
=
33
100

故答案为:
33
100