数列{an}的前n项和为sn,sn=12n2+12n,则数列{1anan+1}的前100项的和为()A.100101B.99101C.99100D.101100-数学

题目简介

数列{an}的前n项和为sn,sn=12n2+12n,则数列{1anan+1}的前100项的和为()A.100101B.99101C.99100D.101100-数学

题目详情

数列{an}的前n项和为snsn=
1
2
n2+
1
2
n
,则数列{
1
anan+1
}
的前100项的和为(  )
A.
100
101
B.
99
101
C.
99
100
D.
101
100
题型:单选题难度:中档来源:不详

答案

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

当n=1时,a1=s1=1
当n≥2时,an=sn-sn-1=class="stub"1
2
n2+class="stub"1
2
n-class="stub"1
2
(n-1)2-class="stub"1
2
(n-1)
=n
而n=1时,a1=1适合上式
故an=n
∴S100=1-class="stub"1
2
+class="stub"1
2
-class="stub"1
3
+…+class="stub"1
100
-class="stub"1
101
=1-class="stub"1
101
=class="stub"100
101

故选A

更多内容推荐