问题
解答题
| 已知等差数列{an}的首项a1=3，公差d≠0，其前n项和为Sn，且a1，a4，a13成等比数列． （Ⅰ）求数列{an}的通项公式； （Ⅱ）证明：
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答案
由a1，a4，a13成等比数列，得a42=a1a13，
即(a1+3d)2=a1(a1+12d)，所以a12+6a1d+9d2=a12+12a1d．
9d2=6a1d，a1=
| 3 |
| 2 |
| 2 |
| 3 |
| 2 |
| 3 |
（Ⅰ）an=a1+（n-1）d=3+2（n-1）=2n+1；
（Ⅱ）Sn=na1+
| n(n-1)d |
| 2 |
则
| 1 |
| Sn |
| 1 |
| n(n+2) |
| 1 |
| 2 |
| 1 |
| n |
| 1 |
| n+2 |
所以
| 1 |
| S1 |
| 1 |
| S2 |
| 1 |
| Sn |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 3 |
| 1 |
| 5 |
| 1 |
| n-1 |
| 1 |
| n+1 |
| 1 |
| n |
| 1 |
| n+2 |
=
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| n+1 |
| 1 |
| n+2 |
| 3 |
| 4 |
| 1 |
| 2(n+1) |
| 1 |
| 2(n+2) |
| 3 |
| 4 |