Question

下列哪一个表达式等价于 $\begin{aligned} &(2 + 3)(2^2 + 3^2)(2^4 + 3^4) \\ &\quad {}\cdot (2^8 + 3^8)(2^{16} + 3^{16}) \\ &\quad {}\cdot (2^{32} + 3^{32})(2^{64} + 3^{64})? \end{aligned}$

Which of the following is equivalent to $\begin{aligned} &(2 + 3)(2^2 + 3^2)(2^4 + 3^4) \\ &\quad {}\cdot (2^8 + 3^8)(2^{16} + 3^{16}) \\ &\quad {}\cdot (2^{32} + 3^{32})(2^{64} + 3^{64})? \end{aligned}$

$3^{127} + 2^{127}$

$3^{127} + 2^{127} + 2 \cdot 3^{63} + 3 \cdot 2^{63}$

$3^{128} - 2^{128}$

$3^{128} + 2^{128}$

$5^{127}$

答案：C

知识点：平方差裂项相消

难度评级：1560

解答：

因为 $3 - 2 = 1$ ，将乘积乘以 $3 - 2$ 不改变它的值。于是再乘以下一个因子 $(3^2 + 2^2)$ 得到 $3^4 - 2^4$ ，依此类推。每一步都会使指数加倍。 $(3-2)(3+2) = 3^2 - 2^2,$

用完全部七个因子后，乘积望远镜化为 $3^{128} - 2^{128}$ 。

因此，正确答案是 C。

Since $3 - 2 = 1,$ multiplying the product by $3 - 2$ does not change it. Then $(3-2)(3+2) = 3^2 - 2^2,$ and multiplying by the next factor $(3^2 + 2^2)$ gives $3^4 - 2^4,$ and so on. Each step doubles the exponent.

After using all seven factors, the product telescopes to $3^{128} - 2^{128}.$

Thus, the correct answer is C.

2021 AMC 12A Spring 第 9 题

答案：C

难度评级：1560

解答：

其他年份的第 9 题