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\(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+...+\frac{2}{899}\)
\(=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+...+\frac{2}{29\cdot31}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{29}-\frac{1}{31}\)
\(=\frac{1}{3}-\frac{1}{31}\)
\(=\frac{28}{93}\)
\(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{899}\)
= \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{29.31}\)
= \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{29}-\frac{1}{31}\)
= \(\frac{1}{3}-\frac{1}{31}+0+0+...+0\)
= \(\frac{29}{93}\)
Ta có \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
Ta có \(B=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(=\frac{1}{2}-\frac{1}{7}\)
\(=\frac{5}{14}\)
Ta có \(C=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)
\(=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)
\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(=\frac{1}{6}-\frac{1}{22}\)
\(=\frac{4}{33}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(B=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(B=\frac{1}{2}-\frac{1}{7}\)
\(B=\frac{5}{14}\)
\(C=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)
\(C=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(C=\frac{1}{6}-\frac{1}{22}=\frac{4}{33}\)
Bài 1 mik học xong quên hết òi (mấy bài kia là hok biết luôn :V)
\(A=\frac{1}{2}+\frac{1}{2.3}+..+\frac{1}{2017.2018}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(A=1-\frac{1}{2018}\)
\(A=\frac{2018}{2018}-\frac{1}{2018}\)
\(A=\frac{2017}{2018}\)
hok tốt!!
bài A: áp dụng công thức: 1 + 2 + 3 + ... + n = n x (n + 1) : 2 tính được 5050
bài B: áp dụng công thức: \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\) rồi triệt tiêu gần hết, qui đồng mẫu số tính được B = 99/100
A = 1 + 2 + 3 + 4 + 5 + ... + 99 + 100
= ( 100 + 1 ) x 100 : 2 = 5050
Vậy A = 5050
\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Vậy \(B=\frac{99}{100}\)
Học tốt #
\(A=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}...\frac{9^2}{9.10}\)
\(A=\frac{1.1.2.2.3.3...9.9}{1.2.2.3.3.4...9.10}\)
\(A=\frac{1}{10}\)
\(B=\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(B=\frac{1}{99}-\left(\frac{1}{99.98}+\frac{1}{98.97}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
\(B=\frac{1}{99}-\left(\frac{1}{99}-\frac{1}{98}+\frac{1}{98}-\frac{1}{97}+...+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-1\right)\)
\(B=\frac{1}{99}-\left(\frac{1}{99}-1\right)\)
\(B=\frac{1}{99}-\frac{1}{99}+1\)
\(B=1\)
