2021 AMC 10A Problem 10

Problem 10 of 25EasierAlgebra

Which of the following is equivalent to

\begin{aligned} &(2+3)(2^2+3^2)\\ &\quad\cdot(2^4+3^4)(2^8+3^8)\\ &\quad\cdot(2^{16}+3^{16})(2^{32}+3^{32})\\ &\quad\cdot(2^{64}+3^{64})? \end{aligned}

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Solution

Multiply the product by

3-2=1.

Repeatedly applying the difference-of-squares identity gives

\begin{aligned} (3-2)(3+2)&=3^2-2^2,\\ (3^2-2^2)(3^2+2^2)&=3^4-2^4, \end{aligned}

and the same cancellation continues through the final factor. Therefore the product is

3^{128}-2^{128}.

Thus, C is the correct answer.

Tagged: difference of squares · telescoping