\(\dfrac{2}{1\times2\times3}+\dfrac{2}{2\times3\times4}+\dfrac{2}{3\times4\times5}+...+\dfrac{2}{48\times49\times50}\)
K
Khách
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Những câu hỏi liên quan
17 tháng 10 2017
\(E=\dfrac{11.3^{29}-3^{2^{15}}}{2.3^{14}.2.3^{14}}\)
\(=\dfrac{11.3-3^{30}}{2^2}=\dfrac{33-3^{30}}{4}\)
AD
23 tháng 7 2023
\(1,\\ =\dfrac{2-1}{1\times2}+\dfrac{3-2}{2\times3}+\dfrac{4-3}{3\times4}+\dfrac{5-4}{4\times5}+.....+\dfrac{99-98}{98\times99}+\dfrac{100-99}{99\times100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{100-1}{100}=\dfrac{99}{100}\)
\(2,=\dfrac{13-11}{11\times13}+\dfrac{15-13}{13\times15}+....+\dfrac{21-19}{19\times21}+\dfrac{23-21}{21\times23}\\ =\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+....+\dfrac{1}{19}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{23}\\ =\dfrac{1}{11}-\dfrac{1}{23}\\ =\dfrac{23-11}{11\times23}=\dfrac{12}{253}\)
@seven
23 tháng 7 2023
a: 1/1*2+1/2*3+...+1/99*100
=1-1/2+1/2-1/3+...+1/99-1/100
=1-1/100
=99/100
b: 2/11*13+2/13*15+...+2/21*23
=1/11-1/13+1/13-1/15+...+1/21-1/23
=1/11-1/23
=12/253
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 6 2019
Áp dụng \(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}\left(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(M=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{48.49.50}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{49.50}\right)=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2450}\right)=\frac{306}{1225}\)
19 tháng 7 2018
\(B=\dfrac{5}{1.2}+\dfrac{13}{2.3}+\dfrac{25}{3.4}+\dfrac{41}{4.5}+...+\dfrac{181}{9.10}\)
\(=\left(\dfrac{1}{1.2}+\dfrac{4}{1.2}\right)+\left(\dfrac{1}{2.3}+\dfrac{12}{2.3}\right)+\left(\dfrac{1}{3.4}+\dfrac{24}{3.4}\right)+...+\left(\dfrac{1}{9.10}+\dfrac{180}{9.10}\right)\)
\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}\right)+\left(\dfrac{4}{1.2}+\dfrac{12}{2.3}+...+\dfrac{180}{9.10}\right)\)
\(=\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)+\left(2+2+...+2\right)\)
\(=1-\dfrac{1}{10}+\left(2.9\right)\)
\(=1-\dfrac{1}{10}+18\)
\(=\dfrac{9}{10}+18\)
\(=18\dfrac{9}{10}\)
TN
2
12 tháng 3 2023
D=1/1-1/2+1/2-1/3+1/3-1/4+.....+1/2022-1/2023
=1-1/2023=2022/2023
TP
16 tháng 8 2017
Ta có:
\(D=1.2+2.3+3.4+4.5+...+99.100\)
\(\Leftrightarrow3D=1.2.\left(3-0\right)+2.3+\left(4-1\right)+3.4+\left(5-2\right)+4.5.\left(6-3\right)+...+99.100.\left(101-98\right)\)
\(\Leftrightarrow3D=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+99.100.101-98.99.100\)
\(\Leftrightarrow3D=99.100.101\Leftrightarrow D=\frac{99.100.101}{3}=333300\)
\(B=1.3+2.4+3.5+4.6+...+99.101\)
\(\Leftrightarrow B=\left(1.3+3.5+...+99.101\right)+\left(2.4+4.6+...+98.100\right)\)
\(\Leftrightarrow6B=\left(1.3.\left(5-\left(-1\right)\right)+3.5.\left(7-1\right)+...+99.101.\left(103-97\right)\right)+\left(2.4.\left(6-0\right)+4.6.\left(8-2\right)+...+98.100.\left(102-96\right)\right)\)
\(\Leftrightarrow B=\frac{99.101.103+3}{6}+\frac{98.100.102}{6}=338250\)
Vì các bước gần tương tự như bài a) nên mình bỏ bước.
TP
16 tháng 8 2017
\(C=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\)
\(\Leftrightarrow C=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\right)\)
\(\Leftrightarrow C=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)
\(\Leftrightarrow C=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{49.50}\right)=\frac{1}{2}.\frac{612}{1225}=\frac{306}{1225}\)
K
29 tháng 3 2018
\(\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{99.100}\)
\(=2.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\)
\(=2.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=2.\left(1-\dfrac{1}{100}\right)=2.\dfrac{99}{100}=\dfrac{99}{50}\)
28 tháng 5 2022
a: \(=\left(\dfrac{1}{15}+\dfrac{14}{15}\right)+\left(\dfrac{9}{10}-2-\dfrac{11}{9}\right)+\dfrac{1}{157}\)
\(=1+\dfrac{1}{157}+\dfrac{81-180-110}{90}\)
\(=\dfrac{158}{157}+\dfrac{-209}{90}\simeq-1.315\)
b: \(=\dfrac{1}{5}+\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{2}{6}\)
=1/3-1/3
=0
c: \(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2015\cdot2017}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\)
=2016/2017
\(\dfrac{2}{1\times2\times3}+\dfrac{2}{2\times3\times4}+\dfrac{2}{3\times4\times5}+...+\dfrac{2}{48\times49\times50}\)
\(=\dfrac{1}{1\times2}-\dfrac{1}{2\times3}+\dfrac{1}{2\times3}-\dfrac{1}{3\times4}+\dfrac{1}{3\times4}-\dfrac{1}{4\times5}+...+\dfrac{1}{48\times49}-\dfrac{1}{49\times50}\)
\(=\dfrac{1}{1\times2}-\dfrac{1}{49\times50}\)
\(=\dfrac{1}{2}-\dfrac{1}{2450}\)
\(=\dfrac{612}{1225}\)
\(\text{#}Toru\)
1/2 - 1/49 x50