题目内容
数列{an}满足a1=1，an+1=
（n∈N*）．
（Ⅰ）证明：数列{
}是等差数列；
（Ⅱ）求数列{an}的通项公式an；
（Ⅲ）设bn=
an，求数列{bn}的前n项和Sn．
| 2n+1an |
| an+2n |
（Ⅰ）证明：数列{
| 2n |
| an |
（Ⅱ）求数列{an}的通项公式an；
（Ⅲ）设bn=
| 1 |
| n•2n+1 |
（Ⅰ）由已知可知
=
，即
=
+1，即
-
=1
∴数列{
}是公差为1的等差数列．
（Ⅱ）由（Ⅰ）知
=
+(n-1)×1=2+(n-1)×1=n+1，∴an=
．
（Ⅲ）由（Ⅱ）知bn=
an=
×
∴bn=
=
(
-
)
∴Sn=b1+b2+…+bn=
(1-
)+
(
-
)+…+
(
)=
[(1-
)+(
-
)+…+(
-
)]=
．
| an+1 |
| 2n+1 |
| an |
| an+2n |
| 2n+1 |
| an+1 |
| 2n |
| an |
| 2n+1 |
| an+1 |
| 2n |
| an |
∴数列{
| 2n |
| an |
（Ⅱ）由（Ⅰ）知
| 2n |
| an |
| 2 |
| a1 |
| 2n |
| n+1 |
（Ⅲ）由（Ⅱ）知bn=
| 1 |
| n•2n+1 |
| 1 |
| n•2n+1 |
| 2n |
| n+1 |
∴bn=
| 1 |
| 2n(n+1) |
| 1 |
| 2 |
| 1 |
| n |
| 1 |
| n+1 |
∴Sn=b1+b2+…+bn=
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 2 |
| 1 |
| n |
| 1 |
| n+1 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| n |
| 1 |
| n+1 |
| n |
| 2(n+1) |
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