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\(E=25\left[3\cdot\left(5+4^2+4^3+...+4^{2021}\right)+1\right]\)
\(=25\cdot\left(4^2+4^2+4^3+...+4^{2021}\right)\)
\(=25\cdot4^{2022}⋮4^{2022}\)
Em xem lại đề nhé! Có xuất hiện dấu + không? Hay chỉ là dấu x
Lời giải:
$a=1+5+5^2+5^3+...+5^{2022}+5^{2023}$
$5a=5+5^2+5^3+5^4+....+5^{2023}+5^{2024}$
$\Rightarrow 5a-a=5^{2024}-1$
$\Rightarrow 4a=5^{2024}-1$
$\Rightarrow 4a+1=5^{2024}\vdots 5^{2023}$ (đpcm)
Lời giải:
$A=1+4+4^2+4^3+...+4^{2023}$
$A=1+4+(4^2+4^3+4^4)+(4^5+4^6+4^7)+...+(4^{2021}+4^{2022}+4^{2023})$
$=5+4^2(1+4+4^2)+4^5(1+4+4^2)+....+4^{2021}(1+4+4^2)$
$=5+(1+4+4^2)(4^2+4^5+...+4^{2021})$
$=5+21(4^2+4^5+....+4^{2021})$
Do đó biểu thức chia 21 dư 5
\(A=\dfrac{2024^{2023}+1}{2024^{2024}+1}\)
\(2024A=\dfrac{2024^{2024}+2024}{2024^{2024}+1}=\dfrac{\left(2024^{2024}+1\right)+2023}{2024^{2024}+1}=\dfrac{2024^{2024}+1}{2024^{2024}+1}+\dfrac{2023}{2024^{2024}+1}=1+\dfrac{2023}{2024^{2024}+1}\)
\(B=\dfrac{2024^{2022}+1}{2024^{2023}+1}\)
\(2024B=\dfrac{2024^{2023}+2024}{2024^{2023}+1}=\dfrac{\left(2024^{2023}+1\right)+2023}{2024^{2023}+1}=\dfrac{2024^{2023}+1}{2024^{2023}+1}+\dfrac{2023}{2024^{2023}+1}=1+\dfrac{2023}{2024^{2023}+1}\)
Vì \(2024>2023=>2024^{2024}>2024^{2023}\)
\(=>2024^{2024}+1>2024^{2023}+1\)
\(=>\dfrac{2023}{2024^{2023}+1}>\dfrac{2023}{2024^{2024}+1}\)
\(=>A< B\)
\(#PaooNqoccc\)