Question

объясните, как решать подобное

Answer 1

$(2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)+1=\\ =\dfrac{(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)+(2-1)}{(2-1)} =\\ =(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)+1=\\ =(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)+1=\\ =(2^8-1)(2^8+1)(2^{16}+1)(2^{32}+1)+1=\\=(2^{16}-1)(2^{16}+1)(2^{32}+1)+1=\\ =(2^{32}-1)(2^{32}+1)+1=\\ =2^{64}-1+1=\\ =2^{64}$