[x]为x的整数部分.当n≥2时,则[112+122+132+…+1n2]的值为()A.0B.1C.2D.3-数学

题目简介

[x]为x的整数部分.当n≥2时,则[112+122+132+…+1n2]的值为()A.0B.1C.2D.3-数学

题目详情

[x]为x的整数部分.当n≥2时,则[
1
12
+
1
22
+
1
32
+…+
1
n2
]
的值为(  )
A.0B.1C.2D.3
题型:单选题难度:偏易来源:不详

答案

class="stub"1
12
+class="stub"1
22
+class="stub"1
32
+…+class="stub"1
n2
≤1+class="stub"1
1×2
+class="stub"1
2×3
+…+class="stub"1
(n-1)n

=1+(1-class="stub"1
2
+class="stub"1
2
-class="stub"1
3
+…+class="stub"1
n-1
-class="stub"1
n

=1+(1-class="stub"1
n
)=2-class="stub"1
n

class="stub"1
12
+class="stub"1
22
+class="stub"1
32
+…+class="stub"1
n2
class="stub"1
1×2
+class="stub"1
2×3
+class="stub"1
3×4
+…+class="stub"1
n(n+1)

=1-class="stub"1
2
+class="stub"1
2
-class="stub"1
3
+class="stub"1
3
-class="stub"1
4
+…+class="stub"1
n
-class="stub"1
n+1

=1-class="stub"1
n+1

∴当n≥2时,则[class="stub"1
12
+class="stub"1
22
+class="stub"1
32
+…+class="stub"1
n2
]
=1.
故选B.

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