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\(A=\dfrac{1}{1\times4}+\dfrac{1}{4\times7}+\dfrac{1}{7\times10}+...+\dfrac{1}{100\times103}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+...+\dfrac{3}{100\times103}\right)\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{103}\right)=\dfrac{1}{3}.\dfrac{102}{103}=\dfrac{34}{103}\)
1/1*4 + 1/4*7 + 1/7*10 + ... + 1/97*100
= 1/3(3/1*4 + 3/4*7 + 3/7*10 + ... + 3/97*100)
= 1/3(1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + .... + 1/97 - 1/100)
= 1/3(1 - 1/100)
= 1/3*99/100
= 33/100
\(\frac{11}{1.4}+\frac{11}{4.7}+...+\frac{11}{100.103}\)
\(=\frac{11}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)
\(=\frac{11}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(=\frac{11}{3}\left(1-\frac{1}{103}\right)\)
Tự tính
\(\frac{11}{1.4}+\frac{11}{4.7}+...+\frac{11}{100.103}\)
= \(\frac{11}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)
= \(\frac{11}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)
= \(\frac{11}{3}.\left(1-\frac{1}{103}\right)\)
= \(\frac{11}{3}.\frac{102}{103}\)
= \(\frac{374}{103}\)
Đặt \(B=\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+......+\frac{2}{100\cdot103}\)
\(B=\frac{2}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.....+\frac{1}{100}-\frac{1}{103}\right)\)
\(B=\frac{2}{3}\cdot\left(1-\frac{1}{103}\right)\)
\(B=\frac{2}{3}\cdot\frac{102}{103}\)
\(\Rightarrow B=\frac{68}{103}\)
Đặt \(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{100.103}\)
\(A=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(A=\frac{2}{3}\left(1-\frac{1}{103}\right)\)
\(A=\frac{2}{3}\cdot\frac{102}{103}\)
\(A=\frac{68}{103}\)
\(A=\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+...+\dfrac{3}{100\cdot103}\)
\(=\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{100}-\dfrac{1}{103}\)
\(=\dfrac{98}{515}\)
Ta có: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2020\cdot2021}+\dfrac{1}{2021\cdot2022}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2020}-\dfrac{1}{2021}+\dfrac{1}{2021}-\dfrac{1}{2022}\)
\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)
1/1x2+1/2x3+1/3x4+...+1/2020x2021+1/2021x2022
=1/1-1/2+1/2-1/3+1/3-1/4+...+1/2020-1/2021+1/2021-1/2022.
=1/1-1/2022
=2021/2022
S=1/1-1/4+1/4+1/7-1/7+1/10+...+1/100-1/103
S=1/1-1/103
S=102/103
Vì 102/103<1 nên S<1
Có công thức \(\dfrac{x}{a\left(a+x\right)}=\dfrac{1}{a}-\dfrac{1}{a+x}\) nhé!
Ví dụ: \(\dfrac{2}{2.4}=\dfrac{1}{2}-\dfrac{1}{4}\)
\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
\(=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\)
\(=1-\dfrac{1}{8}=\dfrac{7}{8}\)
Dấu . tức là nhân nhé!
\(F=\dfrac{1}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{100\cdot103}\right)\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{102}{103}=\dfrac{34}{103}\)