问题
解答题
设an=
|
答案
∵an=
+1•2
+…+2•3
>1+2+…+n=n(n+1)
(5分)n(n+1) 2
又∵an=
+1•2
+…+2•3 n(n+1)
<
+1+2 2
+…+2+3 2 n(n+1) 2
=
=n(n+1)+n(N+3) 4
<n2+2n 2
(11分)(n+1)2 2
∴
<an<n(n+1) 2
(12分)(n+1)2 2
设an=
|
∵an=
+1•2
+…+2•3
>1+2+…+n=n(n+1)
(5分)n(n+1) 2
又∵an=
+1•2
+…+2•3 n(n+1)
<
+1+2 2
+…+2+3 2 n(n+1) 2
=
=n(n+1)+n(N+3) 4
<n2+2n 2
(11分)(n+1)2 2
∴
<an<n(n+1) 2
(12分)(n+1)2 2