122-1+132-1+142-1+…+1(n+1)2-1的值为()A.n+12(n+2)B.34-n+12(n+2)C.34-12(1n+1+1n+2)D.32-1n+1-1n+2-数学

题目简介

122-1+132-1+142-1+…+1(n+1)2-1的值为()A.n+12(n+2)B.34-n+12(n+2)C.34-12(1n+1+1n+2)D.32-1n+1-1n+2-数学

题目详情

1
22-1
+
1
32-1
+
1
42-1
+…+
1
(n+1)2-1
的值为(  )
A.
n+1
2(n+2)
B.
3
4
-
n+1
2(n+2)
C.
3
4
-
1
2
(
1
n+1
+
1
n+2
)
D.
3
2
-
1
n+1
-
1
n+2
题型:单选题难度:偏易来源:不详

答案

class="stub"1
22-1
+class="stub"1
32-1
+class="stub"1
42-1
+…+class="stub"1
(n+1)2-1

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

=class="stub"1
3×1
+class="stub"1
4×2
+class="stub"1
5×3
+…+class="stub"1
(n+2)n

=class="stub"1
2
[(1-class="stub"1
3
)+(class="stub"1
2
-class="stub"1
4
)  +(class="stub"1
3
-class="stub"1
5
)+…+ (class="stub"1
n
-class="stub"1
n+2
)
+(class="stub"1
2
-class="stub"1
4
)
+(class="stub"1
3
-class="stub"1
5
)
+…+(class="stub"1
n
-class="stub"1
n+2
)
]
=class="stub"1
2
(1+class="stub"1
2
-class="stub"1
n+1
-class="stub"1
n+2
)

=class="stub"3
4
-class="stub"1
2
(class="stub"1
n+1
+class="stub"1
n+2
)

故选C.

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