问题
填空题
数列{an}的通项公式an=n(n+1),Sn为数列{
|
答案
∵
=1 an
=1 n(n+1)
-1 n
,1 n+1
∴Sn=(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
数列{an}的通项公式an=n(n+1),Sn为数列{
|
∵
=1 an
=1 n(n+1)
-1 n
,1 n+1
∴Sn=(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