Rút gọn: \(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+......+\frac{1}{20}\left(1+2+3+...+20\right)\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right)..................\left(1-\frac{1}{20}\right)\)
=\(\frac{1}{2}.\frac{2}{3}.............\frac{19}{20}\)
=\(\frac{1.2.3..............19}{2.3.4..............20}\)
=\(\frac{1}{20}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}......\frac{21}{20}\)
\(B=\frac{21}{2}\)
@@@
\(B=\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{20}\right)\)
\(\Rightarrow B=\left(\frac{2}{2}+\frac{1}{2}\right)\left(\frac{3}{3}+\frac{1}{3}\right)\left(\frac{4}{4}+\frac{1}{4}\right)...\left(\frac{20}{20}+\frac{1}{20}\right)\)
\(\Rightarrow B=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{21}{20}\)
\(\Rightarrow B=\frac{21}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
B=\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{19}{20}\)
B=\(\frac{1.2.3....19}{2.3.4....20}\)
B=\(\frac{1}{20}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{20}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{19}{20}=\frac{1}{20}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{20}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{19}{20}\)
\(=\frac{1\cdot2\cdot...\cdot19}{2\cdot3\cdot...\cdot20}\)
\(=\frac{1}{20}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\left[1-\frac{1}{2}\right]\cdot\left[1-\frac{1}{3}\right]\cdot\left[1-\frac{1}{4}\right]\cdot...\cdot\left[1-\frac{1}{20}\right]\)
\(B=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{19}{20}\)
\(B=\frac{1\cdot2\cdot3\cdot...\cdot19}{2\cdot3\cdot4\cdot...\cdot20}=\frac{1}{20}\)
Lời giải:
$A=1+\frac{2.3}{2.2}+\frac{3.4}{2.3}+\frac{4.5}{2.4}+...+\frac{20.21}{2.20}$
$=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{21}{2}$
$=\frac{1+2+3+...+21}{2}-\frac{1}{2}=\frac{21.22}{2.2}-\frac{1}{2}=115$