Question

18 So sánh 2 số: A=3^32-1 và B=(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)

Answer 1

\(B=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(2B=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(2B=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(2B=\left(3^{16}-1\right)\left(3^{16}+1\right)\)

\(B=\dfrac{3^{32}-1}{2}< A=3^{32}-1\)

Answer 2

\(B=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =>2B=2.\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^{16}-1\right)\left(3^{16}+1\right)=3^{32}-1\\ =>A=\dfrac{3^{32}-1}{2}< B\)

Answer 3

\(B=\left(3+1\right)\left(3^2+1\right)...\left(3^{16}+1\right)=\dfrac{\left(3-1\right)\left(3+1\right)...\left(3^{16}+1\right)}{2}=\dfrac{\left(3^2-1\right)\left(3^2+1\right)...\left(3^{16}+1\right)}{2}=\dfrac{\left(3^4-1\right)\left(3^4+1\right)...\left(3^{16}+1\right)}{2}=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^8+16\right)}{2}=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{2}=\dfrac{3^{32}-1}{2}=\dfrac{A}{2}\)

Vậy \(A=2B\)