Tính:
\(D=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{\left(5n+1\right).\left(5n+6\right)}\)
Tính đầy đủ hộ mik vs. Mik đag cần gấp
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IM
30 tháng 8 2016
Ta có
\(\frac{1}{1.6}+\frac{1}{6.11}+......+\frac{1}{\left(5n+1\right)\left(5n+6\right)}\)
\(=\frac{1}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+.....+\frac{1}{5n+1}-\frac{1}{5n+6}\right)\)
\(=\frac{1}{5}\left(1-\frac{1}{5n+6}\right)\)
\(=\frac{1}{5}.\left[\frac{\left(5n+6\right)-1}{\left(5n+6\right)}\right]\)
\(=\frac{1}{5}.\frac{5n+5}{5n+6}\)
\(=\frac{n+1}{5n+6}\)
\(\Rightarrow\frac{1}{1.6}+\frac{1}{6.11}+......+\frac{1}{\left(5n+1\right)\left(5n+6\right)}=\frac{n+1}{5n+6}\) ( đpcm )
1 tháng 3 2022
\(VT=\dfrac{1}{5}\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{\left(5n+1\right)\left(5n+6\right)}\right)\)
\(=\dfrac{1}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-...+\dfrac{1}{5n+1}-\dfrac{1}{5n+6}\right)\)
\(=\dfrac{1}{5}\left(1-\dfrac{1}{5n+6}\right)\)
\(=\dfrac{1}{5}\cdot\dfrac{5n+6-1}{5n+6}\)
\(=\dfrac{n+1}{5n+6}=VP\)
MT
14 tháng 7 2015
\(B=\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right)...\left(1-\frac{1}{10000}\right)\)
\(=\left(\frac{4}{4}-\frac{1}{4}\right).\left(\frac{9}{9}-\frac{1}{9}\right)...\left(\frac{10000}{10000}-\frac{1}{10000}\right)\)
\(=\frac{3}{4}.\frac{8}{9}...\frac{9999}{10000}=\frac{3}{2.2}.\frac{2.4}{3.3}...\frac{99.101}{100.100}\)
\(=\frac{101}{100}\)
\(D=\frac{5^2}{1.6}+\frac{5^2}{6.11}+\frac{5^2}{11.16}+...+\frac{5^2}{26.31}\)
\(=5.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{26.31}\right)\)
\(=5.\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{26}-\frac{1}{31}\right)\)
\(=5.\left(\frac{1}{1}-\frac{1}{31}\right)=5.\left(\frac{31}{31}-\frac{1}{31}\right)=5.\frac{30}{31}=\frac{150}{31}\)
BM
21 tháng 3 2016
Ta có :
\(\frac{5}{1.6}+\frac{5}{6.11}+................+\frac{5}{\left(5.x+1\right).\left(5.x+6\right)}=\)\(\frac{50}{41}\)
=> \(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...............+\frac{1}{5.x+1}-\frac{1}{5.x+6}\) = \(\frac{50}{41}\)
=> \(1-\frac{1}{5.x+6}=\frac{50}{41}\)
=> \(\frac{1}{5.x+6}=\frac{-9}{41}\)................ mình ko tìm ra vì p/s kia ko có tử là 1
bạn xem lại đề bài giúp mình nha
9 tháng 8 2015
Đặt A = \(\frac{1}{1.6}+\frac{1}{6.11}+..+\frac{1}{\left(5n+1\right)\left(5n+6\right)}\)
5A = \(\frac{5}{1.6}+\frac{5}{6.11}+..+\frac{5}{\left(5n+1\right)\left(5n+6\right)}\)
= \(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+..+\frac{1}{5n+1}-\frac{1}{5n+6}\)
= \(\frac{1}{1}-\frac{1}{5n+6}=\frac{5n+6-1}{5n+6}=\frac{5n+5}{5n+6}=\frac{5\left(n+1\right)}{5n+6}\)
=> A = \(=\frac{5\left(n+1\right)}{5n+6}:5=\frac{5\left(n+1\right)}{5n+6}\cdot\frac{1}{5}=\frac{n+1}{5n+6}\)
VẬy VT = VP ĐT Đ CM
D = \(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{\left(5n+1\right)\left(5n+6\right)}\)
= \(\frac{1}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{5n+1}-\frac{1}{5n+6}\right)\)
= \(\frac{1}{5}\left(1-\frac{1}{5n+6}\right)\)
= \(\frac{1}{5}.\frac{5n+5}{5n+6}\)
= \(\frac{n+1}{5n+6}\)