Rút gọn:
\(C=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{\left(3n+2\right)\left(3n+5\right)}\)
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TP
22 tháng 8 2017
Ta có:
\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{\left(3n+2\right).\left(3n+5\right)}\)
\(\Rightarrow\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{\left(3n+2\right).\left(3n+5\right)}\right)\)
\(\Rightarrow\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n+2}-\frac{1}{3n+5}\right)\)
\(\Rightarrow\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{3n+5}\right)\)
\(\Rightarrow\frac{1}{6}-\frac{1}{9n+15}\)
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) .
HA
28 tháng 1 2016
Đặt A=1/2.5+1/5.8+...+1/(3n-1).(3n+2)
=>3A=3/2.5+3/5.8+...+3/(3n-1).(3n+2)
=>3A=1/2-1/5+1/5-1/8+...+1/3n-1-1/3n+2
=>3A=1/2-1/3n+2
=>3A=(3n+2-2)/[2.(3n+2)]
=>3A=3n/6n+4
=>A=3n/6n+4/3
=>A=n/6n+4
14 tháng 4 2016
Đặt \(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+......+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(=>3A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+....+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\)
=> \(3A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+....+\frac{1}{3n-1}-\frac{1}{3n+2}\)
=>\(3A=\frac{1}{2}-\frac{1}{3n+2}\)
=> \(3A=\frac{\left(3n+2\right):2}{3n+2}-\frac{1}{3n+2}\)
=> \(3A=\frac{1,5.n}{3n+2}\)
=>\(A=\frac{1,5.n}{3n+2}.\frac{1}{3}=>A=\frac{1,5.n}{\left(3n+2\right).3}=\frac{1,5.n}{9n+6}\)
\(Hay\) \(A=\frac{1,5n:1,5}{\left(9n+6\right):1,5}=\frac{n}{9n:1,5+6:1,5}=\frac{n}{6n + 4} \left(đpcm\right)\)
24 tháng 7 2019
\(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{3n+2}\right)\)
\(=\frac{1}{3}.\frac{3n}{2.\left(3n+2\right)}\)
\(=\frac{n}{2\left(3n+2\right)}\)
1 tháng 6 2018
Bài 1:
\(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{\left(2x-2\right).2x}\)\(=\frac{11}{48}\)
\(\frac{1}{4}.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(x-1\right).x}\right)\)\(=\frac{11}{48}\)
\(\frac{1}{4}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x-1}-\frac{1}{x}\right)\)\(=\frac{11}{48}\)
\(\frac{1}{4.}.\left(1-\frac{1}{x}\right)=\frac{11}{48}\)
\(1-\frac{1}{x}=\frac{11}{48}:\frac{1}{4}\)
\(1-\frac{1}{x}=\frac{11}{12}\)
\(\frac{1}{x}=1-\frac{11}{12}\)
\(\frac{1}{x}=\frac{1}{12}\)
Vậy x= 12
Bài 2 :
Xét vế trái ta có :
\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{\left(3n-1\right).\left(3n+2\right)}\)
\(=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\right)\)
\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{3n+2}\right)\)
\(=\frac{1}{3}.\frac{1}{2\left(3n+2\right)}=\frac{n}{2\left(3n+2\right)}\)
VẾ TRÁI ĐÚNG BẰNG VẾ PHẢI .ĐẲNG THỨC ĐÃ CHỨNG TỎ LÀ ĐÚNG
cHÚC BẠN HỌC TỐT ( -_- )
\(C=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{\left(3n+2\right)\left(3n+5\right)}\)
\(=\frac{1}{3}\left[\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{\left(3n+2\right)\left(3n+5\right)}\right]\)
\(=\frac{1}{3}\left[\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+...+\frac{\left(3n+5\right)-\left(3n+2\right)}{\left(3n+2\right)\left(3n+5\right)}\right]\)
\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{3n+2}-\frac{1}{3n+5}\right]\)
\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{3n+5}\right]\)
\(=\frac{1}{3}.\frac{3n+5-2}{2\left(3n+5\right)}=\frac{3n+3}{3.2\left(3n+5\right)}=\frac{n+1}{2\left(3n+5\right)}\)