问题 解答题
设a=
1+
1
12
+
1
22
+
1+
1
22
+
1
32
+
1+
1
32
+
1
42
+…+
1+
1
20002
+
1
20012
,问与a最接近的整数是多少?
答案

∵n为任意的正整数,

1+
1
n2
+
1
(n+1)2
=
n2(n+1)2+n2+(n+1)2
[n(n+1)]2

=

[n(n+1)]2+2n(n+1)+1
[n(n+1)]2
=
(n2+n+1)2
[n(n+1)]2
=
n2+n+1
n(n+1)
=1+
1
n(n+1)

∴a=(1+

1
1×2
)+(1+
1
2×3
)+(1+
1
3×4
)+…+(1+
1
2000×2001

=2000+

1
1×2
+
1
2×3
+
1
3×4
+…+
1
2000×2001

=2000+(1-

1
2
)+(
1
2
-
1
3
)+(
1
3
-
1
4
)+…+(
1
2000
-
1
2001
)=2001-
1
2001

因此,与a最接近的整数是2001.

判断题
单项选择题