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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{2}S=\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{21}\)
\(\Rightarrow\left(\frac{1}{2}S\right)-S=\left(\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{21}\right)-\left(\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{20}\right)\)
\(\Rightarrow-\frac{1}{2}S=\left(\frac{1}{2}\right)^{21}-\left(\frac{1}{2}\right)\)
\(\Rightarrow S=\frac{\left(\left(\frac{1}{2}\right)^{21}-\frac{1}{2}\right)}{-\frac{1}{2}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{20}\)(1)
\(\Rightarrow2A=1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{19}\)(2)
Lấy (2) trừ đi (1) ta có :
\(\Rightarrow2A-A=1-\left(\frac{1}{2}\right)^{20}\)
\(\Rightarrow A=1-\left(\frac{1}{2}\right)^{20}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: 1+2+3+...+n=\(\frac{n\left(n+1\right)}{2}\)
=> \(1=\frac{1x2}{2};\frac{1}{2}\left(1+2\right)=\frac{2x3}{2x2};\frac{1}{3}\left(1+2+3\right)=\frac{3x4}{2x3};\)\(;\frac{1}{4}\left(1+2+3+4\right)=\frac{4x5}{2x4};...;\frac{1}{20}\left(1+2+3+...+20\right)=\frac{20x21}{2x20}\)
=> \(B=\frac{1x2}{2}+\frac{2x3}{2x2}+\frac{3x4}{2x3}+\frac{4x5}{2x4}+...+\frac{20x21}{2x20}\)
=> \(B=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{21}{2}\)
=> \(B=\frac{1}{2}\left(2+3+4+5+...+21\right)=\frac{1}{2}\left(\frac{21.22}{2}-1\right)\)
=> \(B=\frac{230}{2}=115\)
Đáp số: B=115
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{n+1}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{n}{n+1}\)
\(=\frac{1}{n+1}\)
\(1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)...+\frac{1}{20}.\left(1+2+3+...+20\right)\)
\(=1+\frac{1}{2}.2.3:2+\frac{1}{3}.3.4:2+\frac{1}{4}.4.5:2+...+\frac{1}{20}.20.21:2\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{21}{2}\)
\(=\frac{2+3+4+5+...+21}{2}=115\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đáp án: thiếu đề
@#@
mời bn xem xét lại đề bài.
~hok tốt~
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Rightarrow B=1+\frac{1}{2}.\frac{\left(1+2\right).2}{2}+\frac{1}{3}.\frac{\left(1+3\right).3}{2}+....+\frac{1}{20}.\frac{\left(1+20\right).20}{2}\)
\(\Rightarrow B=1+\frac{1}{2}.\frac{3.2}{2}+\frac{1}{3}.\frac{4.3}{2}+...+\frac{1}{20}.\frac{21.20}{2}\)
\(\Rightarrow B=1+\frac{1}{2}.3+\frac{4}{2}+...+\frac{21}{2}\)
\(\Rightarrow B=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{21}{2}\)
\(\Rightarrow B=\frac{2+3+4+...+21}{2}=...\)
Good Clever
\(B=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{20}\left(1+2+3+...+20\right)\)
\(=1+\frac{1}{2}\cdot\frac{2\cdot3}{2}+\frac{1}{3}\cdot\frac{3\cdot4}{2}+...+\frac{1}{20}\cdot\frac{20\cdot21}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{21}{2}\)
\(=\frac{1+2+3+....+21}{2}-\frac{1}{2}\)
\(=\frac{21\cdot22}{2}\cdot\frac{1}{2}-\frac{1}{2}\)
\(=\frac{1}{2}\left(\frac{21\cdot22}{2}-1\right)\)
\(=230\cdot\frac{1}{2}\)
Bí
Đặt \(A=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+...+\left(\frac{1}{2}\right)^{20}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{20}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{19}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{19}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{20}}\right)\)
\(A=1-\frac{1}{2^{20}}\)
\(A=\frac{2^{20}}{2^{20}}-\frac{1}{2^{20}}\)
\(A=\frac{2^{20}-1}{2^{20}}\)
Vậy chọn câu a)