1/3+1/6+1/10+.......+1/120
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8 tháng 8 2017
\(C=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...........+\dfrac{1}{120}\)
\(\Leftrightarrow C=\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+..........+\dfrac{2}{240}\)
\(\Leftrightarrow C=2\left(\dfrac{1}{6}+\dfrac{1}{12}+...........+\dfrac{1}{240}\right)\)
\(\Leftrightarrow C=2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+............+\dfrac{1}{15.16}\right)\)
\(\Leftrightarrow C=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.........+\dfrac{1}{15}-\dfrac{1}{16}\right)\)
\(\Leftrightarrow C=2\left(\dfrac{1}{2}-\dfrac{1}{16}\right)\)
\(\Leftrightarrow C=2.\dfrac{7}{16}=\dfrac{7}{8}\)
8 tháng 8 2017
\(C=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{120}\)
\(\dfrac{1}{2}C=\dfrac{1}{2}\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{120}\right)\)
\(\dfrac{1}{2}C=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{240}\)
\(\dfrac{1}{2}C=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{15}-\dfrac{1}{16}\)
\(\dfrac{1}{2}C=\dfrac{1}{2}-\dfrac{1}{16}\)
\(\dfrac{1}{2}C=\dfrac{7}{16}\)
\(C=\dfrac{7}{8}\)
NH
0
LQ
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
20 tháng 12 2023
A = \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + \(\dfrac{1}{15}\) +.. . + \(\dfrac{1}{120}\)
A = \(\dfrac{2}{2}\).(\(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + \(\dfrac{1}{15}\) + ... + \(\dfrac{1}{120}\))
A = 2.( \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\) + \(\dfrac{1}{30}\) + ... + \(\dfrac{1}{240}\))
A = 2.( \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + ... + \(\dfrac{1}{15.16}\))
A =2 .( \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{15}\) - \(\dfrac{1}{16}\))
A = 2.( \(\dfrac{1}{2}\) - \(\dfrac{1}{16}\))
A = 2.\(\dfrac{7}{16}\)
A = \(\dfrac{7}{8}\)
NH
2
KT
9 tháng 8 2018
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+....+\frac{1}{105}+\frac{1}{210}\)
=> \(\frac{1}{2}A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+.....+\frac{1}{210}+\frac{1}{240}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+.....+\frac{1}{14.15}+\frac{1}{15.16}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{!}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{16}\)
\(=\frac{1}{2}-\frac{1}{16}=\frac{7}{16}\)
=> \(A=\frac{7}{8}\)
TN
1
7 tháng 3 2015
A=2.1/2.3+2.1/2.6+2.1/2.10+....+2.1/2.120
A=2(1/6+1/12+1/20+...+1/240)
A=2(1/2.3+1/3.4+1/4.5+...+1/15.16)
Vì 1/2.3+1/3.4+1/4.5+...+1/15.16=(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+...+(1/15-1/16)=1/2-1/16=7/16
Nên A=2.7/16=7/8
LV
0
NT
1
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{120}\)
\(=2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{240}\right)\)
\(=2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{15\cdot16}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{16}\right)\)
\(=2\left(\frac{8}{16}-\frac{1}{16}\right)\)
\(=2\cdot\frac{7}{16}\)
\(=\frac{7}{8}\)
\(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{120}\)
\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{240}\)
\(=2\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{240}\right)\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{15.16}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{16}\right)\)
\(=\frac{7}{8}\)