Question

Bài 5: So sánh hai số P và Q biết :
P = 4(3^2 + 1)(3^4 + 1) …(3^64 + 1) và Q = 3^218 – 1

Answer 1

\(P=4\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)=\dfrac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)=\dfrac{1}{2}\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)=\dfrac{1}{2}\left(3^{128}-1\right)< 3^{128}-1=Q\)

Answer 2

\(P=4\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\\ 2P=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\\ 2P=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\\ 2P=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\\ 2P=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)...\left(3^{64}+1\right)\\ 2P=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\left(3^{64}+1\right)\\ 2P=\left(3^{32}-1\right)\left(3^{32}+1\right)\left(3^{64}+1\right)\\ 2P=\left(3^{64}-1\right)\left(3^{64}+1\right)=3^{128}-1\\ P=\dfrac{3^{128}-1}{2}< Q=3^{218}-1\)