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Lời giải:
Ta có:
\(A=\frac{(2^3+1)(3^3+1)(4^3+1)...(100^3+1)}{(2^3-1)(3^3-1).....(100^3-1)}\)
\(=\frac{(2+1)(2^2-2+1)(3+1)(3^2-3+1).....(100+1)(100^2-100+1)}{(2-1)(2^2+2+1)(3-1)(3^2+3+1)...(100-1)(100^2+100+1)}\)
\(=\frac{3.4...101(2^2-2+1)(3^2-3+1)...(100^2-100+1)}{1.2.3..99(2^2+2+1)(3^2+3+1)...(100^2+100+1)}\)
\(=\frac{100.101}{2}.\frac{(2^2-2+1)(3^2-3+1)....(100^2-100+1)}{(2^2+2+1)(3^2+3+1)...(100^2+100+1)}\)
Xét: \(a^2+a+1=(a+1)^2-a=(a+1)^2-(a+1)+1\)
Do đó:
\(\left\{\begin{matrix} 2^2+2+1=3^2-3+1\\ 3^2+3+1=4^2-4+1\\ ....\\ 99^2+99+1=100^2-100+1\\ \end{matrix}\right.\)
\(\Rightarrow A=\frac{100.101}{2}.\frac{2^2-2+1}{100^2+100+1}=5050.\frac{3}{10101}\)
\(A< 5050.\frac{3}{10100}=\frac{5050}{10100}.3=\frac{3}{2}\)
Vậy \(A< \frac{3}{2}\) hay \(A< B\)
a/ \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(\Rightarrow2A=3^{128}-1\Rightarrow A=\dfrac{3^{128}-1}{2}\)
Baì này mình mới làm lúc sáng bạn vào câu hỏi tương tự có đấy
\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow2A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(\Rightarrow2A=3^{32}-1\)
\(\Rightarrow A=\frac{3^{32}-1}{2}< 3^{32}-1=C\)
