2/2.4+2/4.6+2/6.8+.....+2/n.(n+2)=49/100
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25 tháng 2 2022
\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{98.100}\\ =\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{98}-\dfrac{1}{100}\\ =\dfrac{1}{2}-\dfrac{1}{100}\\ =\dfrac{49}{100}\)
21 tháng 1 2022
\(=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{50}-\dfrac{1}{52}=\dfrac{1}{2}-\dfrac{1}{52}=\dfrac{25}{52}\)
25 tháng 7 2023
Bài 1 :
\(S=1.3+3.5+5.7+...+99.101=3+15+35+...9999\)
Ta thấy :
\(3=2^2-1\)
\(15=4^2-1\)
\(35=6^2-1\)
.....
\(9999=100^2-1\)
\(\Rightarrow S=2^2+4^2+...+100^2-\left(1\right).\left(\left(100-2\right):2+1\right)\)
\(\Rightarrow S=\dfrac{100.\left(100+1\right)\left(2.100+1\right)}{6}-51\)
\(\Rightarrow S=\dfrac{100.101.201}{6}-51=338299\)
17 tháng 8 2016
\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2004.2006}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2004}-\frac{1}{2006}\)
\(=\frac{1}{2}-\frac{1}{2006}\)
\(=\frac{1003}{2006}-\frac{1}{2006}\)
\(=\frac{1002}{2006}\)
\(=\frac{501}{1003}\)
TM
3
15 tháng 9 2023
a. \(\dfrac{1}{2.4}\) + \(\dfrac{1}{4.6}\) + \(\dfrac{1}{6.8}\) + ...... + \(\dfrac{1}{20.22}\)
= 1/2 ( 1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + ..... + 1/20 - 1/22)
=1/2 ( 1/2 - 1/22)
= 1/2 . 5/11
= 5/22
b. 1+ 2/3 + 2/6 + 2/10 +...+ 2/45
=>1/2.(1+2/3+2/6+....+2/45)=1/2+2/6+2/12+...+2/90
=1/2+2/2.3+2/3.4+...+2/9.10
=2.(1/4+3-2/2.3+4-3/3.4+...+10-9/9.10)
=2. ( 1/4+1/2-1/3+1/3-1/4+.....+1/9-1/10)
= 2.( 1/4-1/10)=2.3/20=3/10
=> vì 1/2.*=3/10
=> *=3/10:1/2=3/5
tick mình nhé
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
16 tháng 9 2023
B = 1 + \(\dfrac{2}{3}\) + \(\dfrac{2}{6}\) +\(\dfrac{2}{10}\) + \(\dfrac{2}{15}\)+...+ \(\dfrac{2}{45}\)
B = 1 + 2.(\(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + \(\dfrac{1}{15}\)+...+ \(\dfrac{1}{45}\))
B = 1 + \(\dfrac{4}{2}\).(\(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) +\(\dfrac{1}{10}\) + \(\dfrac{1}{15}\) + ...+ \(\dfrac{1}{45}\))
B = 1 + 4.( \(\dfrac{1}{6}\) +\(\dfrac{1}{12}\)+ \(\dfrac{1}{20}\)+ \(\dfrac{1}{30}\)+...+ \(\dfrac{1}{90}\))
B = 1 + 4.(\(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\)+ \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\)+...+\(\dfrac{1}{9.10}\))
B = 1 + 4 .( \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\)+...+ \(\dfrac{1}{9}\) - \(\dfrac{1}{10}\))
B = 1 + 4.( \(\dfrac{1}{2}\) - \(\dfrac{1}{10}\))
B = 1 + 4. \(\dfrac{2}{5}\)
B = \(\dfrac{13}{5}\)
NT
1
29 tháng 4 2021
Sửa đề: \(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2018\cdot2020}+\dfrac{4}{2020\cdot2022}\)
Ta có: \(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2018\cdot2020}+\dfrac{4}{2020\cdot2022}\)
\(=2\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2018\cdot2020}+\dfrac{2}{2020\cdot2022}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2018}-\dfrac{1}{2020}+\dfrac{1}{2020}-\dfrac{1}{2022}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{2022}\right)\)
\(=2\cdot\dfrac{505}{1011}\)
\(=\dfrac{1010}{1011}\)
RT
2
11 tháng 7 2021
a) Ta có: \(\dfrac{1}{2022}-\dfrac{5}{2\cdot4}-\dfrac{5}{4\cdot6}-\dfrac{5}{6\cdot8}-...-\dfrac{5}{2020\cdot2022}\)
\(=\dfrac{1}{2022}-5\left(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{2020\cdot2022}\right)\)
\(=\dfrac{1}{2022}-\dfrac{5}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2020\cdot2022}\right)\)
\(=\dfrac{1}{2022}-\dfrac{5}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2020}-\dfrac{1}{2022}\right)\)
\(=\dfrac{1}{2022}-\dfrac{5}{2}\left(\dfrac{1}{2}-\dfrac{1}{2022}\right)\)
\(=\dfrac{1}{2022}-\dfrac{5}{2}\cdot\dfrac{1010}{2022}\)
\(=\dfrac{1}{2022}-\dfrac{2025}{2022}=\dfrac{-1262}{1011}\)
11 tháng 7 2021
b) Ta có: \(\dfrac{2^2}{1\cdot3}+\dfrac{2^2}{3\cdot5}+...+\dfrac{2^2}{197\cdot199}\)
\(=2\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{197\cdot199}\right)\)
\(=2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{197}-\dfrac{1}{199}\right)\)
\(=2\left(1-\dfrac{1}{199}\right)\)
\(=2\cdot\dfrac{198}{199}=\dfrac{396}{199}\)
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