问题
填空题
已知数列{an}的通项公式an=
|
答案
∵an=
=1+2+…+n n
=n(n+1) 2n
，∴bn=n+1 2
=1 anan+1
=4(4 (n+1)(n+2)
-1 n+1
)，1 n+2
∴数列{bn}的前n项和=4[(
-1 2
)+(1 3
-1 3
)+…+(1 4
-1 n+1
)]1 n+2
=4(
-1 2
)=1 n+2
．2n n+2
故答案为
．2n n+2