2021 AMC 12A Spring 第 9 题

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}\text{？}

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}

把整个乘积乘以

3 - 2 = 1

，启动裂项相消的连锁

Multiply the whole product by

3 - 2 = 1

to start a telescoping cascade

反复使用

(3^k - 2^k)(3^k + 2^k) = 3^{2k} - 2^{2k}

Repeatedly apply

(3^k - 2^k)(3^k + 2^k) = 3^{2k} - 2^{2k}

因为

3 - 2 = 1

，将乘积乘以

3 - 2

不改变它的值。于是

(3-2)(3+2) = 3^2 - 2^2\text{，}

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。

用完全部七个因子后，乘积逐层相消，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.

其他年份的第 9 题