CHo G = 5/3 + 8/32 + 11/33 +...+ 302/3100 Chứng minh 2 5/9(hỗn số) <G< 3 1/2 Nhất định sẽ hậu tạ like ^^
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
3G = 3(5/3+8/3^2+11/3^3+...+302/3^100) = 5 + 8/3 + 11/3^2 + ... + 302/3^99
3G - G = ( 5 + 8/3 + 11/3^2 + ... + 302/3^99 ) - ( 5/3+8/3^2+11/3^3+...+302/3^100 )
2G = 5 + 8/3 + 11/3^2 + ... + 302/3^99 - 5/3 - 8/3^2 - 11/3^3 - ... - 302/3^100
2G = 5 + 1 + 1/3 + 1/3^2 + ... + 1/3^98 - 302/3^100 (1)
Đặt B = 1/3 + 1/3^2 + 1/3^3 + ... + 1/3^98
3B = 1 + 1/3 + 1/3^2 + ... + 1/3^97
3B - B = 1 + 1/3 + 1/3^2 + ... + 1/3^97 - 1/3 - 1/3^2 - 1/3^3 - ... - 1/3^98
2B = 1 - 1/3^98
B = 1/2 - 1/3^98 (2)
Từ (1) và (2) => 2G = 5 + 1 + 1/3 + 1/3^2 + ... + 1/3^98 - 302/3^100 = 6 + 1/2 - 1/3^98
=> G = 3 + 1/4 - 1/3^98
ta có 0 < 1/4 - 1/3^98 < 1/2
=> 3 < 3 + 1/4 - 1/3^98 < 3 + 1/2
suy ra 2 + 5/9 < G < 3 + 1/2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=3+3^2+3^3+3^4+.......+3^{100}\)
\(\Rightarrow A=\left(3+3^2+3^3+3^4\right)+.......+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)
\(\Rightarrow A=3.\left(1+3+3^2+3^3\right)+........+3^{97}.\left(1+3+3^2+3^3\right)\)
\(\Rightarrow A=3.40+.........+3^{97}.40\)
\(\Rightarrow A=40.\left(3+.......+3^{97}\right)\)
\(\Rightarrow A⋮40\)( 1 )
Vì \(A\)là tổng của các bậc lũy thừa của 3 nên \(A⋮3\)( 2 )
Từ ( 1 ) và ( 2 ) suy ra : \(A⋮40.3\)
\(\Rightarrow A⋮120\)
Vậy \(A⋮120\)( ĐPCM )
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
B= 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + 9 + 10 - 11 - 12 +...+ 298 - 299 - 300 + 301 + 302
= 1 + ( 2 - 3 - 4 + 5) + ( 6 - 7 - 8 + 9) + ( 10 - 11 - 12 + 13) +...+ (298 - 299 - 300 + 301 ) + 302
= 1 + 0 + 0 +...+ 0 + 302
= 1 + 302 = 303 chia hết cho 3
=> B chia hết cho 3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=3^0+3^1+3^2...+3^{100}\)
\(=3^0\times\left(1+3^1+3^2\right)+3^3\times\left(1+3^1+3^2\right)+...+3^{98}\times\left(1+3^1+3^2\right)\)
\(=3^0\times13+3^3\times13+...+3^{98}\times13\)
\(=13\times\left(3^0+3^3+...+3^{98}\right)⋮13\)