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chỉnh lại đề: so sánh ..... và 2200
\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{100}+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{100}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{100}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{100}+1\right)\)
\(=\left(2^{100}-1\right)\left(2^{100}+1\right)=2^{200}-1< 2^{200}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có
\(A=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{99}{2^{99}}+\frac{100}{2^{100}}\)
\(2A=1+\frac{2}{2}+\frac{3}{2^2}+...+\frac{99}{2^{98}}+\frac{100}{2^{99}}\)
Suy ra \(A=2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\frac{100}{2^{100}}\)
Đặt \(n=\frac{1}{2}\) thì \(A=1+n+n^2+...+n^{99}-\frac{100}{2^{100}}\)
Xét \(B=1+n+n^2+...+n^{99}\Leftrightarrow B.n=n+n^2+n^3+...+n^{100}\)
\(\Leftrightarrow B.n=\left(1+n+n^2+...+n^{99}\right)+\left(n^{100}-1\right)\)
\(\Leftrightarrow B.n=B+n^{100}-1\Leftrightarrow B\left(n-1\right)=n^{100}-1\Leftrightarrow B=\frac{n^{100}-1}{n-1}\)
Suy ra \(A=\frac{\frac{1}{2^{100}}-1}{\frac{1}{2}-1}-\frac{100}{2^{100}}=2\left(1-\frac{1}{2^{100}}\right)-\frac{100}{2^{100}}=-\frac{102}{2^{100}}+2< 2\)
Vậy A < 2
![](https://rs.olm.vn/images/avt/0.png?1311)
Làm dễ hiểu chút
\(A=\left(2^2+4^2+...+100^2\right)-\left(1^2+3^2+...+99^2\right)\)
\(=\left(2^2-1^2\right)+\left(4^2-3^2\right)+...+\left(100^2-99^2\right)\)
\(=\left(2+1\right)\left(2-1\right)+\left(4+3\right)\left(4-3\right)+...+\left(100-99\right)\left(99+100\right)\)
\(=3+7+...+199\)
\(B=3^8.7^8-\left(21^4-1\right)\left(21^4+1\right)\)
\(=21^8-\left(21^8-1\right)=1\)
Vậy A > B
gọi biểu thức \(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{100}+1\right)\) là A
Ta có:\(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{100}+1\right)\)
\(\Rightarrow A=1+\dfrac{1}{2}+...+\dfrac{1}{2^{100}}\)
\(\Rightarrow2.A=2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{99}}\)
\(\Rightarrow2A-A=\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{99}}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{2^{100}}\right)\)
\(\Rightarrow2-\dfrac{1}{2^{100}}< 2^{100}\)
hay \(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{100}+1\right)< 2^{100}\)
bạn giải thích dòng 2 dòng 3 đi bạn mk ko hiểu