GIÚP MÌNH ZỚI:
Cho S=\(\frac{5}{2.3.4}\)+\(\frac{5}{3.4.5}\)+....+\(\frac{5}{98.99.100}\)+\(\frac{5}{99.100.101}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
10 tháng 5 2016
\(S=\frac{5}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)
\(S=\frac{5}{2}.\left(\frac{1}{2.3}-\frac{1}{100.101}\right)\)
\(S=\frac{5}{2}.\left(\frac{5047}{30300}\right)\Rightarrow S=\frac{5047}{12120}\)
8 tháng 11 2016
A = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{99.100.101}\)
=> A = \(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)
= \(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{100.101}\right)\)
= \(\frac{1}{2}.\frac{5049}{10100}\)
= \(\frac{5049}{20200}\)
8 tháng 11 2016
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{99.100.101}\)
\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{99.100.101}\)
Ta thấy:
\(\frac{2}{1.2.3}=\frac{1}{1.2}-\frac{1}{2.3};\frac{2}{2.3.4}=\frac{1}{2.3}-\frac{1}{3.4};...;\frac{2}{99.100.101}=\frac{1}{99.100}-\frac{1}{100.101}\)
\(\Rightarrow2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\)
\(\Rightarrow2A=\frac{1}{1.2}-\frac{1}{100.101}\)
\(\Rightarrow2A=\frac{1}{2}-\frac{1}{10100}\)
\(\Rightarrow2A=\frac{5050}{10100}-\frac{1}{10100}\)
\(\Rightarrow2A=\frac{5049}{10100}\Rightarrow A=\frac{5049}{10100}:2=\frac{5049}{20200}\)
S
30 tháng 7 2018
\(=\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)-\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{99.100.101}\right)\)
\(=\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)
\(=\left(1-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{100.101}\right)\)
\(=\frac{99}{100}-\frac{1}{2}\cdot\frac{5049}{10100}=\frac{99}{100}-\frac{5049}{20200}=\frac{14949}{20200}\)
HT
Tính:
S = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+\frac{1}{4.5.6}+...+\frac{1}{98.99.100}\)
3
10 tháng 8 2016
\(2S=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(2S=\frac{1}{2}-\frac{1}{9900}\)
\(2S=\frac{4949}{9900}\)
\(S=\frac{4949}{19800}\)
VH
11 tháng 8 2016
Ta xét : \(\frac{1}{1.2}-\frac{1}{2.3}=\frac{2}{1.2.3}\)
\(\frac{1}{2.3}-\frac{1}{3.4}=\frac{2}{2.3.4}\)
...
\(\frac{1}{98.99}-\frac{1}{99.100}=\frac{2}{98.99.100}\)
Ta có : 2S = \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\)
=> 2S = \(\frac{1}{1.2}-\frac{1}{99.100}\)
=> 2S = \(\frac{4949}{9900}\)
=> S = \(\frac{4949}{19800}\)
G
4
17 tháng 10 2016
\(\frac{1}{2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{99.100.101}\)
\(=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{99.100.101}\)
\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{99.100.101}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{100.101}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{10100}\right)\)
\(=\frac{1}{2}.\frac{5049}{10100}=\frac{5049}{20200}\)
17 tháng 10 2016
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{99.100.101}\)
\(\Leftrightarrow A=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)
\(\Leftrightarrow A=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{100.101}\right)\)
\(\Leftrightarrow A=\frac{1}{2}.\frac{5049}{10100}=\frac{5049}{20200}\)
18 tháng 4 2016
\(\frac{2.9.8+3.12.10+4.15.12+...+98.297.200}{2.3.4+3.4.5+4.5.6+...+98.99.100}=\frac{3.2.\left(2.3.4+3.4.5+4.5.6+...+98.99.100\right)}{2.3.4+3.4.5+4.5.6+...+98.99.100}=6\)
NN
1
HH
3 tháng 12 2015
\(S:3.2=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+....+\frac{2}{98.99.100}\)
\(\frac{1}{1.2}-\frac{1}{2.3}=\frac{2}{1.2.3}\)
\(\frac{1}{2.3}-\frac{1}{3.4}=\frac{2}{2.3.4}\)
Tương tự nhé ta có
\(\frac{2S}{3}=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...-\frac{1}{99.100}=\frac{1}{2}-\frac{1}{9900}=\frac{4949}{9900}\)
\(S=\frac{4949}{6600}\)
làm tiếp theo
\(S=\frac{5}{2}.\left(\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}+\frac{2}{99.100.101}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+....+\frac{1}{98.99}-\frac{1}{99.100}+\frac{1}{99.100}-\frac{1}{100.101}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{2.3}-\frac{1}{100.101}\right)\)
còn lại tự làm
\(S=\frac{5}{2\cdot3\cdot4}+\frac{5}{3\cdot4\cdot5}+......+\frac{5}{99\cdot100\cdot101}\)
\(S\frac{2}{5}=\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+.....+\frac{2}{99\cdot100\cdot101}\)
\(\frac{2}{2\cdot3\cdot4}=\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\)
\(\frac{2}{3\cdot4\cdot5}=\frac{1}{3\cdot4}-\frac{1}{4\cdot5}\)
.............
\(\frac{2}{99\cdot100\cdot101}=\frac{1}{99\cdot100}-\frac{1}{100\cdot101}\)
\(\Rightarrow S\frac{2}{5}=\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+.........+\frac{1}{99\cdot100}-\frac{1}{100\cdot101}\)
\(\Rightarrow S\frac{2}{5}=\frac{1}{2\cdot3}-\frac{1}{100\cdot101}\)
\(\Rightarrow S\frac{2}{5}=\frac{1}{6}-\frac{1}{10100}\)
\(\Rightarrow S\frac{2}{5}=\frac{5047}{30300}\)
\(\Rightarrow S=\frac{5047}{30300}:\frac{2}{5}\)
\(\Rightarrow S=\frac{5047}{30300}\cdot\frac{5}{2}\)
\(\Rightarrow S=\frac{5047}{12120}\)