\(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\frac{1}{2}\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\frac{1}{2}\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\frac{1}{2}\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\frac{1}{2}\left(3^{32}-1\right)\left(3^{32}+1\right)\)
\(=\frac{1}{2}\left(3^{64}-1\right)\)
Đặt \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(\Rightarrow\left(3-1\right)A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)...\left(3^{32}+1\right)\)
\(\Rightarrow2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{32}+1\right)\)
\(\Rightarrow2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)...\left(3^{32}+1\right)\)
Cứ làm như thế cuối cùng ta được
\(2A=3^{64}-1\)
\(\Rightarrow A=\frac{3^{64}-1}{2}\)
