AMC 10/12A Spring 2021 (Problem 10)Which of the following is equivalent to (2+3)(22+32)(24+34)(28+38)(216+316)(232+332)(264+364)(2+3)(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})(2^{64}+3^{64})(2+3)(22+32)(24+34)(28+38)(216+316)(232+332)(264+364) ?(A) 3127+2127\text{(A)}\;3^{127}+2^{127}(A)3127+2127(B) 3127+2127+2⋅363+3⋅263\text{(B)}\;3^{127}+2^{127}+2\cdot3^{63}+3\cdot2^{63}(B)3127+2127+2⋅363+3⋅263(C) 3128−2128\text{(C)}\;3^{128}-2^{128}(C)3128−2128(D) 3128+2128\text{(D)}\;3^{128}+2^{128}(D)3128+2128(E) 5127\text{(E)}\;5^{127}(E)5127Related TopicsToolkit 10 — Difference of squaresHints (3)Hint 1Use 10. Difference of Squares.Hint 2Multiply by (3−2)(3-2)(3−2).Hint 3(3−2)(2+3)(22+32)(24+34)⋯(264+364)=(32−22)(22+32)(24+34)⋯=(34−24)(24+34)⋯=(38−28)(28+38)⋯⇒⋯=3128−2128(3-2)(2+3)(2^2+3^2)(2^4+3^4)\cdots(2^{64}+3^{64}) =(3^2-2^2)(2^2+3^2)(2^4+3^4)\cdots =(3^4-2^4)(2^4+3^4)\cdots =(3^8-2^8)(2^8+3^8)\cdots \Rightarrow \cdots = 3^{128}-2^{128}(3−2)(2+3)(22+32)(24+34)⋯(264+364)=(32−22)(22+32)(24+34)⋯=(34−24)(24+34)⋯=(38−28)(28+38)⋯⇒⋯=3128−2128. So the original product is (3128−2128)/(3−2)=3128−2128(3^{128}-2^{128})/(3-2)=3^{128}-2^{128}(3128−2128)/(3−2)=3128−2128.Final Answer(C) 3128−21283^{128}-2^{128}3128−2128