问题
填空题
观察式子
|
答案
原式=
(1-1 2
)+1 3
(1 2
-1 3
)+…+1 5
(1 2
-1 2n-1
)1 2n+1
=
(1-1 2
+1 3
-1 3
+…+1 5
-1 2n-1
)1 2n+1
=
(1-1 2
)1 2n+1
=
×1 2 2n 2n+1
=
.n 2n+1
故答案为
.n 2n+1
观察式子
|
原式=
(1-1 2
)+1 3
(1 2
-1 3
)+…+1 5
(1 2
-1 2n-1
)1 2n+1
=
(1-1 2
+1 3
-1 3
+…+1 5
-1 2n-1
)1 2n+1
=
(1-1 2
)1 2n+1
=
×1 2 2n 2n+1
=
.n 2n+1
故答案为
.n 2n+1