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PT
1
15 tháng 7 2018
Ta có : A = 1/ 2.5 + 1/ 5.8 + 1/ 8.11 + ... + 1/ (3n-1).(3n+2) .
= 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/ 3n-1 - 1/ 3n+2 .
= 1/2 - 1/ 3n+2 .
= 3n + 2 - 2 / 2 .( 3n+2 ) .
= 3n / 2.(3n+2) .
6 tháng 3 2016
đặt \(\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+....+\frac{1}{65.68}\right)\)là A
Ax=\(\frac{19}{68}+\frac{7}{34}=\frac{33}{68}\)
3A=\(3.\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{11}{8.11}+...+\frac{1}{65.68}\right)\)
3A=\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{65.68}\)
3A=\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{65}-\frac{1}{68}\)
3A=\(\frac{1}{2}-\frac{1}{68}=\frac{33}{68}\)
A=33/68:3=11/68
\(\Rightarrow\)33/68:11/68=3
vậy x= 3
TP
22 tháng 8 2017
Ta có:
\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{\left(3n+2\right).\left(3n+5\right)}\)
\(\Rightarrow\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{\left(3n+2\right).\left(3n+5\right)}\right)\)
\(\Rightarrow\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n+2}-\frac{1}{3n+5}\right)\)
\(\Rightarrow\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{3n+5}\right)\)
\(\Rightarrow\frac{1}{6}-\frac{1}{9n+15}\)
HN
4
OY
5 tháng 9 2021
Đặt A=\(\dfrac{1}{2.5}+\dfrac{1}{5.8}+...+\dfrac{1}{95.98}\)
\(3A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{95.98}\)
\(3A=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{95}-\dfrac{1}{98}\)
\(3A=\dfrac{1}{2}-\dfrac{1}{98}\)
\(3A=\dfrac{24}{49}\Rightarrow A=\dfrac{8}{49}\)
5 tháng 9 2021
\(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}+\dfrac{1}{95.98}\)
\(=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{95}-\dfrac{1}{98}\)
\(=\dfrac{1}{2}-\dfrac{1}{98}\)
\(=\dfrac{24}{49}\)
KH
2
25 tháng 9 2016
<=> \(\frac{1}{3}\cdot\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
<=>\(\frac{1}{3}\cdot\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
<=>\(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}\cdot3=\frac{303}{1540}\)
<=>\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)
<=>\(x+3=308\)
<=>\(x=305\)
\(1-\frac{1}{2\cdot5}-\frac{1}{5\cdot8}-\frac{1}{8\cdot11}-...-\frac{1}{92\cdot95}\)
\(=1-\left(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{92\cdot95}\right)\)
\(=1-\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{2}{92\cdot95}\right)\)
\(=1-\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}\right)\)
\(=1-\frac{1}{3}\left(\frac{1}{2}-\frac{1}{95}\right)\)
\(=1-\frac{1}{3}\cdot\frac{93}{190}\)
\(=1-\frac{31}{190}\)
\(=\frac{159}{190}\)
\(1-\frac{1}{2.5}-\frac{1}{5.8}-\frac{1}{8.11}-...-\frac{1}{92.95}\)
\(=1-\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{92.95}\right)\)
\(=1-\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{92.95}\right)\)
\(=1-\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}\right)\)
\(=1-\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{95}\right)\)
\(=1-\frac{1}{3}.\frac{93}{190}\)
\(=1-\frac{31}{190}\)
\(=\frac{159}{190}\)