已知数列{an}:12,13+23,14+24+34,15+25+35+45,…,那么数列{bn}={1anan+1}前n项的和为( ) A.4(1-1n+1) B.4(12-1n+1) C.1-1n+1 D.12-1n+1
问题描述:
已知数列{an}:
,1 2
+1 3
,2 3
+1 4
+2 4
,3 4
+1 5
+2 5
+3 5
,…,那么数列{bn}={4 5
}前n项的和为( )1
anan+1
A. 4(1-
)1 n+1
B. 4(
-1 2
)1 n+1
C. 1-
1 n+1
D.
-1 2
1 n+1
答
数列{an}的通项公式为an=1n+1+2n+1+3n+1+…+nn+1=n(n+1)2(n+1)=n2数列{bn}={1anan+1}的通项公式为bn=1anan+1=2n•2n+1=4(1n-1n+1)其前n项的和为4[(11-12)+(12-13)+(13-14)+…+(1n-1n+1)]=4(1-1n+1)故选A...