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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+...+\frac{3^2}{97.100}\)
\(A=\frac{3^2}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{97}-\frac{1}{100}\right)\)
\(A=3\cdot\left(1-\frac{1}{100}\right)\)
\(A=3\cdot\frac{99}{100}=\frac{297}{100}\)
Vậy \(A=\frac{297}{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(A=\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}\)
=> \(3A=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
=> \(A+3A=1-\frac{1}{3^{100}}\)
=> \(4A=\frac{3^{100}-1}{3^{100}}\)
=> \(A=\frac{3^{100}-1}{4.3^{100}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
3A=3.(1/3-1/3^2+1/3^3-...+1/3^99-1/3^100)
3A+A=(1-1/3+1/3^2+1/3^3-...+1/3^98-1/3^99)+(1/3-1/3^2+1/3^3-...+1/3^99-1/3^100)
4A= 1-1/3+1/3^2+1/3^3-...+1/3^98-1/3^99+1/3-1/3^2+1/3^3-...+1/3^99-1/3^100
4A=1-1/3^100
A=(1-1/3^100):4
chọn câu trả lời nha
![](https://rs.olm.vn/images/avt/0.png?1311)
\(D=1-3+3^2-3^3+3^4-3^5+...+3^{98}-3^{99}+3^{100}\)
\(3D=3\left(1-3+3^2-3^3+3^4-3^5+...+3^{98}-3^{99}+3^{100}\right)\)
\(3D=3-3^2+3^3-3^4+3^5-3^6+...+3^{99}-3^{100}+3^{101}\)
\(3D+D=\left(3-3^2+3^3-3^4+3^5-3^6+...+3^{99}-3^{100}+3^{101}\right)\)
\(+\left(1-3+3^2-3^3+3^4-3^5+...+3^{98}-3^{99}+3^{100}\right)\)
\(4D=3^{101}+1\) \(\Rightarrow D=\frac{3^{101}+1}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=3-3^2+3^3-3^4+...-3^{100}\)
\(3A=3^2-3^3+3^4-3^5+...-3^{101}\)
\(4A=3-3^{101}\)
\(A=\frac{3-3^{101}}{4}\)
Gọi biểu thức trên là A
=> 3A = 3(3 - 32 + 33 - 34 + ... - 3100)
= 32 - 33 + 34 - ... + 3100 - 3101
=> A + 3A = 4A = (3 - 32 + 33 - 34 + ... - 3100) + (32 - 33 + 34 - ... + 3100 - 3101)
= 3 - 3101
=> A = \(\frac{3-3^{101}}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=3+\frac{3}{1+2}+\frac{3}{1+2+3}+\frac{3}{1+2+3+4}+...+\frac{3}{1+2+3+4+...+100}\)
\(B=3.\left(\frac{1}{\left(1+0\right).2:2}+\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+\frac{1}{\left(1+4\right).4:2}+...+\frac{1}{\left(1+100\right).100:2}\right)\)
\(B=3.\left(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{100.101}\right)\)
\(B=6.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(B=6.\left(1-\frac{1}{101}\right)\)
\(B=6.\frac{100}{101}=\frac{600}{101}\)
Đặt \(A=3-3^2+3^3-3^4+....-3^{100}\)
\(\Rightarrow\)\(3A=3^2-3^3+3^4-3^5+.....-3^{101}\)
\(\Rightarrow\)\(A+3A=\left(3-3^2+3^3-3^4+...-3^{100}\right)+\left(3^2-3^3+3^4-3^5+...-3^{101}\right)\)
\(\Rightarrow\)\(4A=3-3^{101}\)
\(\Rightarrow\)\(A=\frac{3-3^{101}}{4}\)