问题
填空题
若an=1+2+3+…+n,则Sn为数列{
|
答案
由题意可得,an=n(n+1) 2
∴
=1 an
=2(2 n(n+1)
-1 n
)1 n+1
∴Sn=a1+a2+…+an
=2(1-
+1 2
-1 2
+…+1 3
-1 n
)1 n+1
=2(1-
)=1 n+1 2n n+1
故答案为:2n n+1
若an=1+2+3+…+n,则Sn为数列{
|
由题意可得,an=n(n+1) 2
∴
=1 an
=2(2 n(n+1)
-1 n
)1 n+1
∴Sn=a1+a2+…+an
=2(1-
+1 2
-1 2
+…+1 3
-1 n
)1 n+1
=2(1-
)=1 n+1 2n n+1
故答案为:2n n+1