问题
填空题
数列1+
|
答案
Sn=1+
+2+1 2
+…+n+1 4 1 2n
=(1+2+…+n)+(
+1 2
+…+1 4
)1 2n
=
+n(n+1) 2
[1-(1 2
)n]1 2 1- 1 2
=
+1-(n(1+n) 2
)n1 2
故答案为：
+1-(n(n+1) 2
)n1 2
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数列1+
|
Sn=1+
+2+1 2
+…+n+1 4 1 2n
=(1+2+…+n)+(
+1 2
+…+1 4
)1 2n
=
+n(n+1) 2
[1-(1 2
)n]1 2 1- 1 2
=
+1-(n(1+n) 2
)n1 2
故答案为：
+1-(n(n+1) 2
)n1 2