数列112,314,518,7116,…,(2n-1)+12n,…,的前n项和Sn的值为()A.n2+1-12nB.2n2-n+1-12nC.n2+1-12n-1D.n2-n+1-12n-数学

题目简介

数列112,314,518,7116,…,(2n-1)+12n,…,的前n项和Sn的值为()A.n2+1-12nB.2n2-n+1-12nC.n2+1-12n-1D.n2-n+1-12n-数学

题目详情

数列1
1
2
3
1
4
5
1
8
7
1
16
,…,(2n-1)+
1
2n
,…,的前n项和Sn的值为(  )
A.n2+1-
1
2n
B.2n2-n+1-
1
2n
C.n2+1-
1
2n-1
D.n2-n+1-
1
2n
题型:单选题难度:偏易来源:不详

答案

由于Sn=1class="stub"1
2
+3class="stub"1
4
+5class="stub"1
8
+7class="stub"1
16
+…+[(2n-1)+class="stub"1
2n
]
=[1+3+5+7+…+(2n-1)]+(class="stub"1
2
+class="stub"1
4
+class="stub"1
8
+…+class="stub"1
2n

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

=n2+1-class="stub"1
2n

Sn=n2+1-class="stub"1
2n

故答案为 A.

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