题目内容
计算:(1+
)×(1+
)×(1+
)×…×(1+
).
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 8 |
| 1 |
| 2n |
考点:平方差公式
专题:计算题
分析:原式变形后,利用平方差公式计算即可得到结果.
解答:解:原式=2×(1-
)×(1+
)×(1+
)×(1+
)×…×(1+
)
=2×(1-
)×(1+
)×(1+
)×…×(1+
)
=2×(1-
)×(1+
)×…×(1+
)
=2×(1-
).
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 8 |
| 1 |
| 2n |
=2×(1-
| 1 |
| 4 |
| 1 |
| 4 |
| 1 |
| 8 |
| 1 |
| 2n |
=2×(1-
| 1 |
| 8 |
| 1 |
| 8 |
| 1 |
| 2n |
=2×(1-
| 1 |
| 22n |
点评:此题考查了平方差公式,熟练掌握平方差公式是解本题的关键.
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