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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{101+100+99+98+...+3+2+1}{101-100+99-98+...+3-2+1}\)
\(=\frac{\frac{101.102}{2}}{51}\)
\(=101\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=100-98+...+4-2(100 -98 =2 là 1 cặp => A có 25 cặp)
=25 . 2 =50
b ( mk hok bt lm )
C=1+2-3-4+5+6-7-8+...-95-96+97+98( 1+2-3-4 là 1 cặp => có 24 cặp dưa 97 và 98)
= (24 . -4) + 97 + 98 =-96 + 97 + 98 =99
T I C K mk nha!!
![](https://rs.olm.vn/images/avt/0.png?1311)
A = SCSH: ( 102 - 1 ) : 1 + 1 = 102
A = Tổng: ( 102 + 1 ) . 102 : 2 = 5253
Vậy KQ là: 5253
B = SCSH: ( 2998 - 1 ) : 3 + 1 = 1000
B = Tổng: ( 2998 + 1 ) . 1000 : 2 = 1499500
Vậy KQ là 1499500
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(S=\frac{1}{3}+\frac{2}{3^2}+.......+\frac{101}{3^{101}}\)
\(\Rightarrow3S=1+\frac{2}{3}+.......+\frac{101}{3^{100}}\)
\(\Rightarrow3S-S=\left(1+\frac{2}{3}+..+\frac{101}{3^{100}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+..+\frac{101}{3^{101}}\right)\)
\(\Rightarrow2S=1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{100}}-\frac{101}{3^{101}}< 1+\frac{1}{3}+....+\frac{1}{3^{100}}\)
\(\Rightarrow6S< 3+1+........+\frac{1}{3^{99}}\)
\(\Rightarrow6S-2S< \left(3+1+....+\frac{1}{3^{99}}\right)-\left(1+\frac{1}{3}+....+\frac{1}{3^{100}}\right)\)
\(\Rightarrow4S< 3-\frac{1}{3^{100}}< 3\Rightarrow S< \frac{3}{4}\)
Đặt \(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}+\frac{101}{3^{101}}\)
\(3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}+\frac{101}{3^{100}}\)
\(3A-A=\left(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{101}{3^{100}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+...+\frac{101}{3^{101}}\right)\)
\(2A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}-\frac{101}{3^{101}}\)
\(6A=3+1+\frac{1}{3}+...+\frac{1}{3^{99}}-\frac{101}{3^{100}}\)
\(6A-2A=\left(3+1+\frac{1}{3}+...+\frac{1}{3^{99}}-\frac{101}{3^{100}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}-\frac{101}{3^{101}}\right)\)
\(4A=3-\frac{101}{3^{100}}-\frac{1}{3^{100}}+\frac{101}{3^{101}}\)
\(4A=3-\frac{303}{3^{101}}-\frac{3}{3^{101}}+\frac{100}{3^{101}}\)
\(4A=3-\frac{206}{3^{101}}< 3\)
=>\(4A< 3\)
\(\Rightarrow A< \frac{3}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1+2-3-4+5+6-7-8+............-99-100+101+102
=1+(2-3-4+5)+(6-7-8+9)+...............+(98-99-100+101)+102
=1+0+0+..............+0+102
=103
![](https://rs.olm.vn/images/avt/0.png?1311)
cho mi sửa lại:
\(a) A = 1^2+2^3+3^4+...+2014^{2015} b) B = 101^2+102^2+...+199^2+200^2 c) C = 1^3+2^4+3^5+4^6+...+99^{101}+100^{102}\)
\(A=\left(2^{101}-2^{100}+2^{99}-2^{98}+....+2^2-2+1\right):\left(2^{102}-1\right)\)
\(S=2^{100}+2^{98}+...+2^2+2^0\text{ do đó: }4S=2^{102}+2^{100}+...+2^4+2^2\Rightarrow4S-S=2^{102}-1\Rightarrow A=S:\left(2^{102}-1\right)=\dfrac{1}{3}\)