Question

So sánh: C=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1) và D=2^32

Answer 1

sửa đề D=2^32-1

ta có:

C=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)

= (2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)

= (2^4-1)(2^4+1)(2^8+1)(2^16+1)

= (2^8-1)(2^8+1)(2^16+1)

=(2^16-1)(2^16-1)

= 2^32-1^2

 

Answer 2

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

\(C=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^6+1\right)\)

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

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

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

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

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

Vì 2³² - 1 < 2³²

=> C < D