\(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
= \(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
= \(\frac{1}{3}-\frac{1}{100}\)
= \(\frac{97}{300}\)
Tính A=\(\frac{1}{2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}\)
\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+.....+\frac{1}{98\cdot99\cdot100}=\frac{1}{k}\cdot\left(\frac{1}{1\cdot2}-\frac{1}{99\cdot100}\right)\)
Số k trong đẳng thức trên có giá trị là ?
\(a=\frac{2\cdot9\cdot8+3\cdot12\cdot10+4\cdot15\cdot12+...+98\cdot297\cdot200}{23\cdot4+3\cdot4\cdot5+4\cdot5\cdot6+...+98\cdot99\cdot100}\)
Tính a2
Tìm y :
\(\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{98\cdot99\cdot100}\right)\cdot y=\frac{49}{200}\)
a) A = \(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot5}+...+\frac{4}{107\cdot111}\)
b) B = \(\frac{6}{15\cdot18}+\frac{6}{18\cdot21}+\frac{6}{21\cdot24}+...+\frac{6}{87\cdot90}\)
c) C = \(\frac{1}{1\cdot6}+\frac{1}{6\cdot11}+\frac{1}{11\cdot16}+...+\frac{1}{51\cdot56}\)
d) D = \(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{98\cdot99\cdot100}\)
Giúp mình với!
CHO \(a=\frac{2\cdot9\cdot8+3\cdot12\cdot10+4\cdot15\cdot12+...+98\cdot297\cdot200}{23\cdot4+3\cdot4\cdot5+4\cdot5\cdot6+...+98\cdot99\cdot100}\)
Tính a2
Tính tổng S= \(\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+.......+\frac{2}{98\cdot99}+\frac{2}{99\cdot100}\)
Ai xong trước mình tích cho
Rút gọn biểu thức:
\(A=\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\frac{4^2}{3\cdot5}......\frac{99^2}{98\cdot100}\)
Chứng minh rằng:\(\frac{7}{12}< \frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{99\cdot100}< \frac{5}{6}\)
