Question

Tính (2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)-2^32

Answer 1

Ta có: \(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)-2^{32}\)

\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)-2^{32}\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)-2^{32}\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)-2^{32}\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)-2^{32}\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)-2^{32}\)

\(=2^{32}-1-2^{32}\)

=-1