题目内容
求证:
+3
+5
+…+(2n+1)
=(n+1)2n.
| C | 0n |
| C | 1n |
| C | 2n |
| C | nn |
证明:设Sn=
+3
+5
+…+(2n+1)
①
把①式右边倒转过来得Sn=(2n+1)
+(2n-1)
+…+3
+
,
又由
=
可得Sn=(2n+1)
+(2n-1)
+…+3
+
②
①+②得 2Sn=(2n+2)(
+
+…+
+
)=2(n+1)•2n,
∴Sn=(n+1)•2n,
即:
+3
+5
+…+(2n+1)
=(n+1)2n,
原等式得证.
| C | 0n |
| C | 1n |
| C | 2n |
| C | nn |
把①式右边倒转过来得Sn=(2n+1)
| C | nn |
| C | n-1n |
| C | 1n |
| C | 0n |
又由
| C | mn |
| C | n-mn |
| C | 0n |
| C | 1n |
| C | n-1n |
| C | nn |
①+②得 2Sn=(2n+2)(
| C | 0n |
| C | 1n |
| C | n-1n |
| C | nn |
∴Sn=(n+1)•2n,
即:
| C | 0n |
| C | 1n |
| C | 2n |
| C | nn |
原等式得证.
练习册系列答案
相关题目