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Bài 1 \(F=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{7.8.9}+\frac{1}{8.9.10}\)
\(2F=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{7.8}-\frac{1}{8.9}+\frac{1}{8.9}-\frac{1}{9.10}\)
\(2F=\frac{1}{1.2}-\frac{1}{9.10}\)\(=\frac{44}{90}\)
\(F=\frac{11}{45}\)
Vậy \(F=\frac{11}{45}\)
Bài 2 :
\(A=\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)
\(\Rightarrow\)\(\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{9.9}\)
\(\Rightarrow\)\(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}< B< \frac{1}{2.3}+..+\frac{1}{8.9}\)
\(\Rightarrow\)\(\frac{1}{3}-\frac{1}{10}< B< \frac{1}{2}-\frac{1}{9}\)
\(\Rightarrow\)\(\frac{7}{30}\)\(< \frac{7}{18}\left(đpcm\right)\)
Hết nha bn.Mk ik ngủ.Chúc bạn học tốt
ta có
tử số \(\frac{1}{19}+\frac{2}{18}+..+\frac{18}{2}+\frac{18}{1}=\frac{1}{19}+1+\frac{2}{18}+1+..+\frac{18}{2}+1\)
\(\frac{20}{19}+\frac{20}{18}+..+\frac{20}{2}=20\left(\frac{1}{19}+\frac{1}{18}+..+\frac{1}{2}\right)\)
Do đó ta có phân số trên bằng 20
\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{20\cdot21\cdot22}=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{20\cdot21\cdot22}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{20\cdot21}-\frac{1}{21\cdot22}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{21\cdot22}\right)\)
\(=\frac{1}{2}\left(\frac{231}{462}-\frac{1}{462}\right)=\frac{1}{2}\cdot\frac{230}{462}=\frac{1}{2}\cdot\frac{115}{231}=\frac{115}{462}\)
\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{2018\cdot2019\cdot2020}\)
\(=\frac{1}{2}\left[\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{2018\cdot2019\cdot2020}\right]\)
\(=\frac{1}{2}\left[\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{2018\cdot2019}-\frac{1}{2019\cdot2020}\right]\)
Đến đây tự tính được rồi:v
Đặt tổng trên là A
Ta có:
\(2A=2\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{2018\cdot2019\cdot2020}\right)\)
\(=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{2018\cdot2019\cdot2020}\)
\(=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{2018\cdot2019}-\frac{1}{2019\cdot2020}\)
\(=\frac{1}{2}-\frac{1}{2019\cdot2020}\)
\(A=\left(\frac{1}{2}-\frac{1}{2019\cdot2020}\right)\div2\)
*Làm tiếp*
\(#Louis\)
* Công thức : \(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{6}\right)=\frac{1}{2}.\left(\frac{3}{6}-\frac{1}{6}\right)=\frac{1}{2}.\frac{2}{6}=\frac{1}{6}=\frac{1}{1.2.3}\)
\(A=\frac{3}{1.2.3}+\frac{3}{2.3.4}+...+\frac{3}{2015.2016.2017}\)
\(\Rightarrow A=3.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2015.2016.2017}\right)\)
\(\Rightarrow A=3.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2015.2016}-\frac{1}{2016.2017}\right)\)
\(\Rightarrow A=3.\left(\frac{1}{1.2}-\frac{1}{2016.2017}\right)\)
\(\Rightarrow A=3.\left(\frac{1}{2}-\frac{1}{4066272}\right)\)
\(\Rightarrow A=3.\left(\frac{2033136}{4066272}-\frac{1}{4066272}\right)\)
\(\Rightarrow A=3.\frac{2033135}{4066272}>3.\frac{1355424}{4066272}\)
\(\Rightarrow A>3.\frac{1}{3}\)
\(\Rightarrow A>1\)
Chúc bạn học tốt !!!
* Công thức :
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{6}\right)=\frac{1}{2}.\left(\frac{3}{6}-\frac{1}{6}\right)=\frac{1}{2}.\frac{2}{6}=\frac{1}{6}=\frac{1}{1.2.3}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{20.21.22}\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{20.21}-\frac{1}{21.22}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{21.22}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{462}\right)\)
\(=\frac{1}{2}.\left(\frac{231}{462}-\frac{1}{462}\right)\)
\(=\frac{1}{2}.\frac{230}{462}\)
\(=\frac{115}{462}\)
Chúc bạn học tốt !!!
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\)
\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{37.38.39}\right)\)
\(=\frac{1}{2}\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{39-37}{37.38.39}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{38.39}\right)\)
\(=\frac{1}{2}.\frac{370}{741}\)
\(=\frac{185}{741}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{48.49.50}\)
= \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{48.49}-\frac{1}{49.50}\)
= \(\frac{1}{1.2}-\frac{1}{49.50}\)
= \(\frac{1}{2}-\frac{1}{2450}\)
= \(\frac{612}{1225}\)
đặt
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{48.49.50}\)
\(\Rightarrow2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{48.49.50}\)
\(\Rightarrow\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{48.49}-\frac{1}{49.50}\)
\(\Rightarrow\frac{1}{1.2}-\frac{1}{49.50}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{2450}=\frac{621}{1225}\)
\(\Rightarrow A=\frac{306}{1225}\)
\(\frac{18}{1.2.3}+\frac{18}{2.3.4}+...+\frac{18}{25.26.27}\)
\(=18.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{25.26.27}\right)\)
\(=18.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{25.26}-\frac{1}{26.27}\right)\)
\(=18.\left(\frac{1}{1.2}-\frac{1}{26.27}\right)\)
\(=18.\left(\frac{1}{2}-\frac{1}{702}\right)\)
\(=18.\left(\frac{351}{702}-\frac{1}{702}\right)\)
\(=18.\frac{350}{702}\)
\(=\frac{350}{39}\)
=18x(1/1x2x3+1/2x3x4+1/3x4x5+...+1/25x26x27). =18x(1/1x2-1/2x3+1/2x3-1/3x4+...+1/25x26-1/26x27). =18x(1/1x2-1/26x27). =18x175/351. =350/39