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Ta có:\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{20}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{19}{20}=\frac{1}{20}\)
\(B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{20}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{19}{20}\)
\(=\frac{1\cdot2\cdot...\cdot19}{2\cdot3\cdot...\cdot20}\)
\(=\frac{1}{20}\)
\(B=\left[1-\frac{1}{2}\right]\cdot\left[1-\frac{1}{3}\right]\cdot\left[1-\frac{1}{4}\right]\cdot...\cdot\left[1-\frac{1}{20}\right]\)
\(B=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{19}{20}\)
\(B=\frac{1\cdot2\cdot3\cdot...\cdot19}{2\cdot3\cdot4\cdot...\cdot20}=\frac{1}{20}\)
B=\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{19}{20}\)
B=\(\frac{1.2.3....19}{2.3.4....20}\)
B=\(\frac{1}{20}\)
B= \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\)\(\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{20}\right)\)
B= \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{19}{20}\)= \(\frac{1}{20}\)
vậy B= \(\frac{1}{20}\)
\(\Rightarrow2A=1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{2014}\)
\(\Rightarrow2A-A=A=1-\left(\frac{1}{2}\right)^{2015}\)
Với B tương tự nhưng là lấy 3B
2.\(B=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.......\frac{49}{50}=\frac{1}{50}\)
\(B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\)
\(=>B=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}=\frac{1\cdot2\cdot3}{2\cdot3\cdot4}=\frac{1}{4}\)
\(B=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}......\frac{21}{20}\)
\(B=\frac{21}{2}\)
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\(B=\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{20}\right)\)
\(\Rightarrow B=\left(\frac{2}{2}+\frac{1}{2}\right)\left(\frac{3}{3}+\frac{1}{3}\right)\left(\frac{4}{4}+\frac{1}{4}\right)...\left(\frac{20}{20}+\frac{1}{20}\right)\)
\(\Rightarrow B=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{21}{20}\)
\(\Rightarrow B=\frac{21}{2}\)