题目内容
已知数列
,
,
,…,
,…,
(1)计算S1,S2,S3,S4;
(2)猜想Sn的表达式,并用数学归纳法证明.
| 1 |
| 1×4 |
| 1 |
| 4×7 |
| 1 |
| 7×10 |
| 1 |
| (3n-2)(3n+1) |
(1)计算S1,S2,S3,S4;
(2)猜想Sn的表达式,并用数学归纳法证明.
(1)S1=
,S2=
,S3=
,S4=
(2)Sn=
证明:①当n=1时,S1=
=
,结论成立
②假设当n=k时成立,结论成立,即Sk=
当n=k+1时,Sk+1=Sk+ ak+1 =
+
=
=
=
∴当n=k+1时结论成立
∴对于任意的k∈N+结论都成立
| 1 |
| 4 |
| 2 |
| 7 |
| 3 |
| 10 |
| 4 |
| 13 |
(2)Sn=
| n |
| 3n+1 |
证明:①当n=1时,S1=
| 1 |
| 3×1+1 |
| 1 |
| 4 |
②假设当n=k时成立,结论成立,即Sk=
| k |
| 3k+1 |
当n=k+1时,Sk+1=Sk+ ak+1 =
| k |
| 3k+1 |
| 1 |
| (3k+1)(3k+4) |
=
| k(3k+4)+1 |
| (3k+1)(3k+4) |
| (k+1)(3k+1) |
| (3k+1)(3k+4) |
| k+1 |
| 3(k+1)+1 |
∴当n=k+1时结论成立
∴对于任意的k∈N+结论都成立
练习册系列答案
相关题目