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
答
∵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
=
(1-1 3
)+1 2
(1 3
-1 2
)+…+1 3
(1 3
-1 99
),1 100
=
(1-1 3
),1 100
=
.33 100
故答案为:
.33 100