已知数列{an}的通项公式an=
|
∵数列{an}的通项公式an=
,1 n(n+1)
∴前n项和Sn=a1+a2+…+an
=
+1 1×2
+…+1 2×3 1 n(n+1)
=(1-
)+(1 2
-1 2
)+…+(1 3
-1 n
)1 n+1
=1-1 n+1
=
.n n+1
故答案为:
.n n+1
已知数列{an}的通项公式an=
|
∵数列{an}的通项公式an=
,1 n(n+1)
∴前n项和Sn=a1+a2+…+an
=
+1 1×2
+…+1 2×3 1 n(n+1)
=(1-
)+(1 2
-1 2
)+…+(1 3
-1 n
)1 n+1
=1-1 n+1
=
.n n+1
故答案为:
.n n+1