3 ×( 2 ²+1) × ( 2^4 + 1 )....( 2^64 + 1) +1

Đáp án:

Giải thích các bước giải: Áp dụng \[\left( {x - 1} \right)\left( {x + 1} \right) = {x^2} - 1\] Ta có \[\begin{array}{l} 3\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right).....\left( {{2^{64}} + 1} \right) + 1\\ = \left( {{2^2} - 1} \right)\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right).....\left( {{2^{64}} + 1} \right) + 1\\ = \left( {{{\left( {{2^2}} \right)}^2} - 1} \right)\left( {{2^4} + 1} \right)......\left( {{2^{64}} + 1} \right) + 1\\ = \left( {{2^4} - 1} \right)\left( {{2^4} + 1} \right)......\left( {{2^{64}} + 1} \right) + 1\\ = \left( {{2^8} - 1} \right)\left( {{2^8} + 1} \right)......\left( {{2^{64}} + 1} \right) + 1\\ .....\\ = \left( {{2^{64}} - 1} \right)\left( {{2^{64}} + 1} \right) + 1\\ = {2^{128}} - 1 + 1 = {2^{128}} \end{array}\]