> 数学 >
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
人气:405 ℃ 时间:2019-11-04 20:34:25
解答
由题意可得Sn=(1+
1
2
)+(3+
1
4
)+(5+
1
8
)+…+(2n-1+
1
2n

=(1+3+5+…+2n-1)+(
1
2
+
1
4
+
1
8
+…+
1
2n

=
n(1+2n−1)
2
+
1
2
(1−
1
2n
)
1−
1
2
=n2+1−
1
2n

故选A
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