\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+.....+\frac{1}{90}\)Tính tổng sau
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)
\(=\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{11.12}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{2}-\frac{1}{12}\)
\(=\frac{5}{12}\)
bn sẽ tinh theo kieeuranhaan 2 nha xin lỗi mik làm bi này rùi nhưng mik quên mik có sacks xem lại
![](https://rs.olm.vn/images/avt/0.png?1311)
a ) Ta có : \(\frac{1}{n}-\frac{1}{n+1}=\frac{n+1}{n.\left(n+1\right)}-\frac{n}{n.\left(n+1\right)}\) \(=\frac{1}{n.\left(n+1\right)}\)
b ) Áp dụng công thức trên tính tổng này như sau :
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
Chúc học giỏi !!!
a, \(VP=\frac{1}{n}-\frac{1}{n+1}=\frac{n+1}{n\left(n+1\right)}-\frac{n}{n\left(n+1\right)}=\frac{n+1-n}{n\left(n+1\right)}\)
\(=\frac{1}{n\left(n+1\right)}=VT\RightarrowĐPCM\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}+\frac{1}{110}\)
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+....+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(=\frac{1}{1}-\frac{1}{11}\)
\(=\frac{10}{11}\)
\(\frac{1}{2}+\frac{1}{3}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}\)\(+\frac{1}{110}\)
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...\) \(+\frac{1}{9\cdot10}\)\(+\frac{1}{10\cdot11}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\)\(\frac{1}{5}\)\(+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}\)\(+\frac{1}{10}-\frac{1}{11}\)
\(=1-\frac{1}{11}\)
\(=\frac{10}{11}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1-\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=1-\left(\frac{1}{90}+\frac{1}{72}+\frac{1}{56}+\frac{1}{42}+\frac{1}{30}+\frac{1}{20}+\frac{1}{12}+\frac{1}{6}+\frac{1}{2}\right)\)
\(=1-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)
\(=1-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
\(=1-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=1-\left(1-\frac{1}{10}\right)\)
\(=1-\frac{9}{10}\)
\(=\frac{1}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Nhớ rằng: \(\frac{n}{a.\left(a+n\right)}=\frac{1}{a}+\frac{1}{a+n}\)
Ta có: \(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
Áp dụng công thức trên, ta có:
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(A=1-\frac{1}{10}=\frac{9}{10}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{1}-\frac{1}{10}\)
\(=\frac{9}{10}\)