chứng tỏ A=1/2+2/22+3/23+...+100/2100<2
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![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{99}+2^{100}\right)\\ A=\left(2+2^2\right)+2^2\left(2+2^2\right)+...+2^{98}\left(2+2^2\right)\\ A=\left(2+2^2\right)\left(1+2^2+...+2^{98}\right)\\ A=6\left(1+2^2+...+2^{98}\right)⋮6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(2+2^2\right)+2^2\left(2+2^2\right)+...+2^{98}\left(2+2^2\right)\)
\(=6+2^2.6+...+2^{98}.6\)
\(=6\left(1+2^2+...+2^{98}\right)⋮6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(2+2^2\right)+2^2\left(2+2^2\right)+...+2^{98}\left(2+2^2\right)\)
\(=6+2^2.6+...+2^{98}.6=6\left(1+2^2+...+2^{98}\right)⋮6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=2+2^2+2^3+...+2^{100}\)
\(=\left(2+2^2\right)+2^2\left(2+2^2\right)+...+2^{98}\left(2+2^2\right)\)
\(=6+6.2^2+...+6.2^{98}\)
\(=6\left(1+2^2+...+2^{98}\right)⋮6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=2+2^2+2^3+2^4+...+2^{100}\)
\(=2\cdot3+2^3\cdot3+...+2^{99}\cdot3\)
\(=6\left(1+2^2+...+2^{98}\right)⋮6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=(2+2^2+2^3+2^4)+2^4(2+2^2+2^3+2^4)+...+2^96(2+2^2+2^3+2^4)
=30(1+2^4+...+2^96) chia hết cho 10
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=2+2^2+2^3+2^4+...+2^{99}+2^{100}\)
\(=\left(2+2^2\right)+2^2\left(2+2^2\right)+...+2^{98}\left(2+2^2\right)\)
\(=6\left(1+2^2+...+2^{98}\right)⋮6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$A=(2+2^2)+(2^3+2^4)+....+(2^{99}+2^{100})$
$=2(1+2)+2^3(1+2)+...+2^{99}(1+2)$
$=2.3+2^3.3+...+2^{99}.3$
$=3(2+2^3+...+2^{99})\vdots 3$
Ta có đpcm.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^2}+....+\frac{100}{2^{100}}\)
\(\Rightarrow2A=1+\frac{2}{2}+\frac{3}{2^2}+...+\frac{100}{2^{99}}\)
\(\Rightarrow2A-A=1+\frac{2}{2}+\frac{3}{2^2}+...+\frac{100}{2^{99}}-\frac{1}{2}-\frac{2}{2^2}-...-\frac{100}{2^{100}}\)
\(\Rightarrow A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}-\frac{100}{2^{100}}\)
\(\Rightarrow A< 1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)(3)
Đặt \(P=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)(2)
\(\Rightarrow2P=2+1+\frac{1}{2}+...+\frac{1}{2^{98}}\)
\(\Rightarrow2P-P=2+1+\frac{1}{2}+...+\frac{1}{2^{98}}-1-\frac{1}{2}-...-\frac{1}{2^{99}}\)
\(\Rightarrow P=2-\frac{1}{2^{99}}< 2\)(1)
Từ (1),(2),(3) => A<2
Giải
Ta có A =1/2 + 2/2^2 + 3/2^3 + ... + 100/2^100
=> 2A = 1 + 2/2 + 3/2^2 + ... + 100/2^99
=> 2A - A = 1 + 1/2 + 1/2^2 + 1/2^3 + ... + 1/2^99 - 100/2^100
=> A = ( 1 - 100/2^100) + 1/2 + 1/2^2 + 1/2^3 + ... + 1/2^99 (*)
Đặt B = 1/2 + 1/2^2 + 1/2^3 + ... + 1/2^99
=> 2B = 1 + 1/2 + 1/2^2 + ... + 1/2^98
=> 2B - B = 1 - 1/2^99
=> B = 1 - 1/2^99
Thay B vào (*) ta được:
A = ( 1 - 100/2^100 ) + ( 1 - 1/2^99 )
A = 2 - ( 100/2^100 + 1/2^99 ) < 2
=> A < 2 (đpcm)