Question

\(\frac{3}{1\cdot2}+\frac{3}{2\cdot3}+\frac{3}{3\cdot4}+................................................+\frac{3}{99\cdot100}\)

Answer 1

Đặt A =  \(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+.....+\frac{3}{99.100}\)

\(\frac{1}{3}A\)\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)

\(\frac{1}{3}A\)\(=1-\frac{1}{100}\)

=> \(\frac{1}{3}A=\frac{99}{100}\)

=> A = \(\frac{99}{100}.3=\frac{297}{100}\)

Answer 2

     \(\frac{3}{1.2}+\frac{3}{2.3}+..................+\frac{3}{99.100}\)

\(=3.\left(\frac{1}{1.2}+\frac{1}{2.3}+..................+\frac{1}{99.100}\right)\)

\(=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.................+\frac{1}{99}-\frac{1}{100}\right)\)

\(=3.\left(1-\frac{1}{100}\right)\)

\(=3.\frac{99}{100}\)

\(=\frac{297}{100}\)

Answer 3

\(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{99.100}\)

\(=\frac{3}{1}-\frac{3}{2}+\frac{3}{2}-\frac{3}{3}+\frac{3}{3}-\frac{3}{4}...+\frac{3}{99}-\frac{3}{100}\)

\(=\frac{3}{1}-\frac{3}{100}\)

\(=\frac{297}{100}\)