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A=\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+..+\frac{1}{9999}\)
\(=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+..+\frac{1}{99.101}\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+..+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\frac{98}{303}=\frac{49}{303}\)
Ta có \(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{998\times999}+\frac{1}{999\times1000}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{998}-\frac{1}{999}+\frac{1}{999}-\frac{1}{1000}\)
\(=1-\frac{1}{1000}\)
\(=\frac{999}{1000}\)
Sửa đề : \(T=\left[1-\frac{1}{4}\right]\times\left[1-\frac{1}{9}\right]\times\left[1-\frac{1}{16}\right]\times...\times\left[1-\frac{1}{576}\right]\times\left[1-\frac{1}{625}\right]\)
\(T=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{575}{576}\cdot\frac{624}{625}\)
\(T=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\frac{3\cdot5}{4\cdot4}\cdot...\cdot\frac{23\cdot25}{24\cdot24}\cdot\frac{24\cdot26}{25\cdot25}\)
\(T=\frac{1\cdot2\cdot3\cdot....\cdot23\cdot24}{2\cdot3\cdot4\cdot....\cdot24\cdot25}\cdot\frac{3\cdot4\cdot5\cdot...\cdot25\cdot26}{2\cdot3\cdot4\cdot...\cdot24\cdot25}\)
Đến đây dễ rồi nhé bạn :v
\(T=\left(1-\frac{1}{4}\right)\times\left(1-\frac{1}{9}\right)\times\left(1-\frac{1}{16}\right)\times...\times\left(1-\frac{1}{157}\right)\times\left(1-\frac{1}{625}\right)\)
\(T=\frac{3}{4}\times\frac{8}{9}\times\frac{15}{16}\times...\times\frac{156}{157}\times\frac{624}{625}\)
\(T=\frac{3\times2\times4\times3\times5\times...\times23\times25\times24\times26}{2\times2\times3\times3\times4\times4\times...\times24\times24\times25\times25}\)
\(T=\frac{1\times26}{2\times25}=\frac{26}{50}=\frac{13}{25}\)
a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)\(=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{98}-\frac{1}{99}\right)+\left(\frac{1}{99}-\frac{1}{100}\right)\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
b) \(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}+\frac{1}{110}\)\(=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}+\frac{1}{10.11}\)
\(=\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{110}\)
\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
c) \(\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+...+\frac{2}{97.99}\) \(=\frac{13-11}{11.13}+\frac{15-13}{13.15}+\frac{17-15}{15.17}+...+\frac{99-97}{97.99}\)
\(=\frac{1}{11}+\frac{1}{13}-\frac{1}{13}+\frac{1}{15}-\frac{1}{15}+\frac{1}{17}...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{11}-\frac{1}{99}=\frac{8}{99}\)
1/1x2 + 1/2x3 + 1/3x4 + 1/4x5 + 1/5x6 + 1/6x7
=1/1-1/2+1/2-1/3+...-1/7
=1+(1/2-1/2+1/3-1/3+...+1/6-1/6)-1/7
=1 +0+0+...-1/7
=1-1/7
=6/7
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+......+1/999-1/1000+1
=1-1/1000+1 (-1/2+1/2=0, -1/3+1/3=0. nên chỉ còn lai các số ko cùng cặp)
=999/1000+1
=1999/1000
Đáp án là 1999/1000
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