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\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{2014.2015.2016}\)
\(A=\dfrac{1}{2}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2014.2015}+\dfrac{1}{2015.2016}\right)\)
\(A=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2015.2016}\right)=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2015.2016}\right)\)
\(A=\dfrac{1}{4}-\dfrac{1}{2.2015.2016}< \dfrac{1}{4}\)
\(=>A< \dfrac{1}{4}\)
Chúc bn học tốt
A= \(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{4.5.6}+....+\dfrac{1}{37.38.39}\)
A=\(\dfrac{1}{1}-\dfrac{1}{39}\)
A=\(\dfrac{38}{39}\)
còn lại tự làm do mình có việc chút
\(S=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{2009.2010.2011}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{2009.2010}-\dfrac{1}{2010.2011}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{2010.2011}\)
\(=\dfrac{1}{2}-\dfrac{1}{4042110}< \dfrac{1}{2}\)
\(\Rightarrow\) \(S< P\)
Vậy \(S< P\)
Gọi biểu thức là \(A\). Ta có :
\(A=\dfrac{3}{1.2.3}+\dfrac{5}{2.3.4}+\dfrac{7}{3.4.5}+...+\dfrac{2017}{1008.1009.1010}\)
\(A=\left(\dfrac{1.2}{1.2.3}+\dfrac{2.2}{2.3.4}+\dfrac{3.2}{3.4.5}+...+\dfrac{1008.2}{1008.1009.1010}\right)+\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{1008.1009.1010}\right)\)\(A=\left(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{1009.1010}\right)+\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{1008.1009}-\dfrac{1}{1009.1010}\right)\)
\(A=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{1009}-\dfrac{1}{1010}\right)+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{1009.1010}\right)\)
\(A< 2.\dfrac{1}{2}+\dfrac{1}{2}.\dfrac{1}{2}=1+\dfrac{1}{4}=\dfrac{5}{4}\)
a) Ta có:
3A= \(1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\left(1\right)\)
A= \(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\left(2\right)\)
Lấy (1) - (2) ta được:
1-\(\dfrac{1}{3^{100}}\)
b) Ta xét:
\(\dfrac{1}{1.2}-\dfrac{1}{2.3}=\dfrac{2}{1.2.3},...,\dfrac{1}{37.38}-\dfrac{1}{38.39}=\dfrac{2}{37.38.39}\)
Ta có:
2B=\(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+..+\dfrac{2}{37.38.39}\)
=\(\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}\right)+\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)+..+\left(\dfrac{1}{37.38}-\dfrac{1}{38.39}\right)\)
=\(\dfrac{1}{1.2}-\dfrac{1}{38.39}=\dfrac{740}{38.39}=\dfrac{370}{741}\)
Vậy \(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+..+\dfrac{2}{37.38.39}\)
=\(\dfrac{370}{741}\)
Nếu bn cảm thấy mk đúng tick cho mk nhé!
\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{18.19.20}\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{18.19}+\dfrac{1}{19.20}\)
\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{18}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{20}\)
\(A=\dfrac{1}{1}-\dfrac{1}{20}\)
\(A=\dfrac{20}{20}-\dfrac{1}{20}\)
\(A=\dfrac{19}{20}\)
A = \(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{18.19.20}\)
A = \(\dfrac{1}{1.2}\)-\(\dfrac{1}{2.3}\)+\(\dfrac{1}{2.3}\)-\(\dfrac{1}{3.4}\)+...+\(\dfrac{1}{18.19}\)-\(\dfrac{1}{19.20}\)
A = \(\dfrac{1}{1.2}\)-\(\dfrac{1}{19.20}\)
A = \(\dfrac{1}{2}\)-\(\dfrac{1}{380}\)
A = \(\dfrac{189}{380}\)
(Mình nghĩ là vậy, có gì sai bạn bỏ qua nha 
)
\(A=\dfrac{1}{1.2}-\dfrac{1}{1.2.3}+\dfrac{1}{2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{3.4}-\dfrac{1}{3.4.5}+\dfrac{1}{99.100}-\dfrac{1}{99.100.101}\)
\(A=\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)-\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{99.100.101}\right)\)
\(A=\left(1-\dfrac{1}{100}\right)-\left(\dfrac{\dfrac{1}{1.2}-\dfrac{1}{100.101}}{2}\right)\)
Bấm máy nha
\(B=\dfrac{5}{1.2.3.4}+\dfrac{5}{2.3.4.5}+\dfrac{5}{3.4.5.6}+...+\dfrac{5}{98.99.100.101}\)
\(B=\dfrac{5}{3}.\left(\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+...+\dfrac{3}{98.99.100.101}\right)\)
\(B=\dfrac{5}{3}.\left(\dfrac{4-1}{1.2.3.4}+\dfrac{5-2}{2.3.4.5}+...+\dfrac{101-98}{98.99.100.101}\right)\)
\(B=\dfrac{5}{3}.\left(\dfrac{4}{1.2.3.4}-\dfrac{1}{1.2.3.4}+\dfrac{5}{2.3.4.5}-\dfrac{2}{2.3.4.5}+...+\dfrac{101}{98.99.100.101}-\dfrac{98}{98.99.100.101}\right)\)
\(B=\dfrac{5}{3}.\left(\dfrac{1}{1.2.3}-\dfrac{1}{99.100.101}\right)\)
\(B=\dfrac{5}{3}.\dfrac{166649}{999900}\approx0,3\)
Đặt S=A
Ta có: \(\frac{3n+2}{n\left(n+1\right)\left(n+2\right)}\)
\(=\frac{2n+2+n}{n\left(n+1\right)\left(n+2\right)}=\frac{2\left(n+1\right)}{n\left(n+1\right)\left(n+2\right)}+\frac{n}{n\left(n+1\right)\left(n+2\right)}\)
\(=\frac{2}{n\left(n+2\right)}+\frac{1}{\left(n+1\right)\left(n+2\right)}=\frac{1}{n}-\frac{1}{n+2}+\frac{1}{n+1}-\frac{1}{n+2}\)
\(=\frac{1}{n}+\frac{1}{n+1}-\frac{2}{n+2}\)
Do đó, ta có: \(\frac{5}{1\cdot2\cdot3}=\frac{3\cdot1+2}{1\cdot2\cdot3}=\frac11+\frac{1}{1+1}-\frac{2}{1+2}=1+\frac12-\frac23\)
\(\frac{8}{2\cdot3\cdot4}=\frac{3\cdot2+2}{2\cdot3\cdot4}=\frac12+\frac13-\frac24\)
...
Do đó, ta có: \(S=1+\frac12-\frac23+\frac12+\frac13-\frac24+\frac13+\frac14-\frac25+\ldots+\frac{1}{n}+\frac{1}{n+1}-\frac{2}{n+2}\)
\(=1+\left(\frac12+\frac12\right)+\left(-\frac23+\frac13+\frac13\right)+\left(-\frac24+\frac14+\frac14\right)+\cdots+\left(-\frac{2}{n}+\frac{1}{n}+\frac{1}{n}\right)-\frac{2}{n+1}+\frac{1}{n+1}-\frac{2}{n+2}\)
\(=1+1-\frac{1}{n+1}-\frac{2}{n+2}<2\)
=>\(S_{2018}<2\)