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\(N=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{1008}+1\right)=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{1008}+1\right)=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{1008}+1\right)=2^{2016}-1< 2^{2016}=M\)
Ta có: \(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1< 2^{32}\)
\(\Leftrightarrow A< B\)
Bài 1: a) \(M=1+5+5^2+...+5^{100}\)
\(5M=5+5^2+5^3+...+5^{101}\)
\(5M-M=\left(5+5^2+5^3+...+5^{101}\right)-\left(1+5+5^2+...+5^{100}\right)\)
\(4M=5^{101}-1\)
\(M=\frac{5^{101}-1}{4}\)
b) \(N=2+2^2+...+2^{100}\)
\(2N=2^2+2^3+...+2^{101}\)
\(2N-N=\left(2^2+2^3+...+2^{101}\right)-\left(2+2^2+...+2^{100}\right)\)
\(N=2^{101}-2\)
Bài 2:
a) \(16^{32}=\left(2^4\right)^{32}=2^{128}\)
\(32^{16}=\left(2^5\right)^{16}=2^{80}\)
Vì \(2^{128}>2^{80}\Rightarrow16^{32}>32^{16}\)
\(N=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)=2^{16}-1< 2^{16}=M\)
\(N=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\\ N=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\\ N=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\\ N=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\\ N=\left(2^8-1\right)\left(2^8+1\right)=2^{16}-1< 2^{16}=M\)
Ta có:
\(N=\left(1+2\right)\left(2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{2008}+1\right)\)
\(\Leftrightarrow N=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{2008}+1\right)\)
\(\Leftrightarrow N=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{2008}+1\right)\)
\(\Leftrightarrow N=\left(2^8-1\right)...\left(2^{2008}+1\right)\)
\(\Leftrightarrow N=2^{4016}-1>2^{2016}=M\)
Ta có :
\(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^{16}+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^{16}+1\right)\)
= ..................................................................
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)=2^{32}-1=B\)
=> A = B
Ta có
N = ( 2 + 1 ) ( 2 2 + 1 ) ( 2 4 + 1 ) ( 2 8 + 1 ) ( 2 16 + 1 ) ( 2 16 + 1 ) = 3 ( 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 16 2 − 1 = 2 32 − 1 M à 2 32 − 1 > 2 32 ⇒ N < M
Đáp án cần chọn là: A