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

题目简介

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

题目详情

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
=______.
题型:填空题难度:中档来源:不详

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

=class="stub"1
3
class="stub"1
2
+class="stub"1
6
+class="stub"1
12
+class="stub"1
20
+…+class="stub"1
100×99
),
=class="stub"1
3
(1-class="stub"1
2
+class="stub"1
2
-class="stub"1
3
+class="stub"1
3
-class="stub"1
4
+class="stub"1
4
-class="stub"1
5
+…+class="stub"1
99
-class="stub"1
100
),
=class="stub"1
3
(1-class="stub"1
100
),
=class="stub"33
100

故答案为:class="stub"33
100

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