Question

rút gọn biểu thức sau A=(2+1)(2^2+1)...(2^64+1)+1

 

Answer 1

help mình nha

 

Answer 2

\(A=\left(2+1\right)\left(2^2+1\right)...\left(2^{64}+1\right)+1\)

\(A=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)...\left(2^{64}+1\right)+1\)

\(A=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)

\(A=2^{128}-1+1=2^{128}\)

Answer 3

Ta có: \(A=\left(2+1\right)\left(2^2+1\right)\cdot...\cdot\left(2^{64}+1\right)+1\)

\(=\left(2^2-1\right)\left(2^2+1\right)\cdot...\cdot\left(2^{64}+1\right)\)+1

\(=\left(2^4-1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{64}+1\right)+1\)

\(=\left(2^8-1\right)\left(2^8+1\right)\cdot...\cdot\left(2^{64}+1\right)+1\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\cdot\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)

\(=\left(2^{32}-1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)

\(=\left(2^{64}-1\right)\left(2^{64}+1\right)+1\)

\(=2^{128}-1+1\)

\(=2^{128}\)