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\(=\frac{2^2-1}{2^2}\cdot\frac{3^2-1}{3^2}\cdot\cdot\cdot\frac{2016^2-1}{2016^2}=\frac{1.3}{2.3}\cdot\frac{2.4}{3.3}\cdot\cdot\cdot\cdot\frac{2015.2017}{2016.2016}\)
\(=\frac{\left(1.2.3....2015\right).\left(3.4....2016.2017\right)}{\left(2.3....2016\right)\left(2.3......2015.2016\right)}=\frac{2017}{2.2016}=\frac{2017}{4032}\)
\(1+\frac{1}{2}.\left(1+2\right)+\)\(\frac{1}{3}.\left(1+2+3\right)+\frac{1}{4}.\left(1+2+3+4\right)+...+\frac{1}{16}.\left(1+2+3+...+16\right)\)
=\(\frac{2}{2}+\frac{3}{2}+\frac{6}{3}+...+\frac{136}{16}\)
=\(\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{17}{2}\)
=\(\frac{2+3+4+5+6+...+17}{2}\)=\(\frac{152}{2}=76\)
Ta có:
\(1+\frac{1}{2}\left(1+2\right)+..........+\frac{1}{20}\left(1+2+3+.......+20\right)\)
\(=1+\frac{1}{2}\left(\frac{3.2}{2}\right)+\frac{1}{3}\left(\frac{4.3}{2}\right)+........+\frac{1}{20}\left(\frac{21.20}{2}\right)\)
\(=1+\frac{3}{2}+\frac{4}{2}+..........+\frac{21}{2}=\frac{2+3+4+........+21}{2}\)
\(=\frac{\frac{23.20}{2}}{2}=\frac{23.10}{2}=115\)
tôi chỉ bn nè muốn làm thì hẳng hok thuộc đề bài vừa hok thuộc vùa nghĩ về bài sẽ nhưng thế nào
Ta có :
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)}=\frac{2003}{2005}\)
\(\Leftrightarrow\)\(1+2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2003}{2005}\)
\(\Leftrightarrow\)\(1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2003}{2005}\)
\(\Leftrightarrow\)\(1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2003}{2005}\)
\(\Leftrightarrow\)\(1+2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2003}{2005}\)
\(\Leftrightarrow\)\(1+1-\frac{2}{x+1}=\frac{2003}{2005}\)
\(\Leftrightarrow\)\(\frac{2}{x+1}=2-\frac{2003}{2005}\)
\(\Leftrightarrow\)\(\frac{2}{x+1}=\frac{2007}{2005}\)
\(\Leftrightarrow\)\(x+1=2:\frac{2007}{2005}\)
\(\Leftrightarrow\)\(x+1=\frac{4010}{2007}\)
\(\Leftrightarrow\)\(x=\frac{4010}{2007}-1\)
\(\Leftrightarrow\)\(x=\frac{2003}{2007}\)
Vậy \(x=\frac{2003}{2007}\)
Chúc bạn học tốt ~
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...........\left(1-\frac{1}{19}\right).\left(1-\frac{1}{20}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.............\frac{18}{19}.\frac{19}{20}\)
\(=\frac{1.2.3...........18.19}{2.3.4...................19.20}=\frac{1.\left(2.3.4..........18.19\right)}{\left(2.3.............19\right).20}=\frac{1}{20}\)
\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{20}\left(1+2+3+...+20\right)=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+...+\frac{1}{20}.\frac{20.21}{2}=1+\frac{3}{2}+\frac{4}{2}+...+\frac{21}{2}=1+\frac{24.19}{2}=229\)
Bài này hơi khó hiểu xíu. Thông cảm nha babe:v
\(B=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+.......+\frac{1}{20}\left(1+2+3+....+20\right)\)
\(B=1+\left(\frac{1}{2}+1\right)+2+\left(\frac{1}{2}+2\right)+3+\left(\frac{1}{2}+3\right)+.....+10+\left(\frac{1}{2}+10\right)\)(chỗ này là nhân phân phối vô đấy!)
\(B=\left(1+2+3+....+10\right)+\left(1+2+3+...+10\right)+\left(\frac{1}{2}.10\right)\)
\(B=55+55+5=115\)
Từ công thức:\(1+2+........+n=\frac{n.\left(n+1\right)}{2}\)
Cho \(n\in\)N*.CMR:\(\frac{1}{n}.\left(1+2+...+n\right)=\frac{n+1}{2}\)
Ta có:\(\frac{1}{n}.\left(1+2+......+n\right)=\frac{1}{n}.\frac{n\left(n+1\right)}{2}=\frac{n+1}{2}\)
Ta có:\(1+\frac{1}{2}\left(1+2\right)+......+\frac{1}{20}.\left(1+2+.....+20\right)\)
\(=1+\frac{1}{2}.\frac{2\left(2+1\right)}{2}+\frac{1}{3}.\frac{3.\left(3+1\right)}{2}+........+\frac{1}{20}.\frac{20\left(20+1\right)}{2}\)
\(=1+\frac{3}{2}+...............+\frac{21}{2}\)
\(=\frac{2+3+......+21}{2}\)
\(=\frac{230}{2}=165\)