问题
填空题
|
答案
∵
=1 n(n+1)
-1 n 1 n+1
∴
+1 1•2
+…1 2•3
=1-1 n(n+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
|
∵
=1 n(n+1)
-1 n 1 n+1
∴
+1 1•2
+…1 2•3
=1-1 n(n+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