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\(S=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\\ 2S=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\\ 2S-S=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right)\\ S=1-\dfrac{1}{2^{100}}=\dfrac{2^{100}-1}{2^{100}}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
mk cũng đang làm bài này, dễ cực luôn
\(B=\frac{5}{28}+\frac{5}{70}+...+\frac{5}{700}\)
\(B=\frac{5}{3}\left[\frac{3}{4.7}-\frac{3}{7.10}+...+\frac{3}{25.28}\right]\)
\(B=\frac{5}{3}\left[\frac{1}{4}-\frac{1}{28}\right]=\frac{5}{14}\)
Chúc bạn học tốt !
![](https://rs.olm.vn/images/avt/0.png?1311)
S = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{100}}\)
2S = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
2S - S = \(1-\frac{1}{2^{100}}\)
=> S = \(1-\frac{1}{2^{100}}\)
Tính :
\(S=2+2^2+2^3+...+2^{100}\)
\(P=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)S=2+22+23+...+2100
2S=2(2+22+23+...+2100)
2S=22+23+...+2101
2S-S=(22+23+...+2101)-(2+22+23+...+2100)
S=2101-2
b)\(P=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\)
\(3P=3\left(\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{100}}\right)\)
\(3P=1+\frac{1}{3}+...+\frac{1}{3^{99}}\)
\(3P-P=\left(1+\frac{1}{3}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)\)
\(2P=1-\frac{1}{3^{100}}\)
\(P=\left(1-\frac{1}{3^{100}}\right):2\)
ngài Kiệt ღ ๖ۣۜLý๖ۣۜ đúng là không ái sánh bằng sự gian xảo này
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(2S=2\left(1+\frac{1}{2}+...+\frac{1}{2^{100}}\right)\)
\(2S=2+1+...+\frac{1}{2^{99}}\)
\(2S-S=\left(2+1+...+\frac{1}{2^{99}}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^{100}}\right)\)
\(S=2-\frac{1}{2^{100}}\)
phần b tương tự
a. S=1+1/2+1/2^2+1/2^3+...+1/2^100
2S=2+1+1/2+1/2^2+...+1/2^99
2S-S=(2+1+1/2+1/2^2+...+1/2^99)-(1+1/2+1/2^2+1/2^3+...+1/2^100)
S=2-1/2^100
S=2^101-1/2^100
\(S=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{100^2}\right)\)
\(S=\frac{3}{2^2}\cdot\frac{8}{3^2}\cdot...\cdot\frac{9999}{100^2}\)
\(S=\frac{\left(1\cdot3\right)\left(2\cdot4\right)\cdot...\cdot\left(99\cdot101\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\cdot...\cdot\left(100\cdot100\right)}\)
\(S=\frac{\left(1\cdot2\cdot...\cdot99\right)\left(3\cdot4\cdot...\cdot101\right)}{\left(2\cdot3\cdot...\cdot100\right)\left(2\cdot3\cdot...\cdot100\right)}\)
\(S=\frac{1\cdot101}{100\cdot2}\)
\(S=\frac{101}{200}\)