Question

2/1×3+2/3×5+2/5×7+...+2/99×101

Answer 1

\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)

\(=1-\frac{1}{101}\)

\(=\frac{100}{101}\)

Answer 2

Ta có : \(\frac{2}{1.3}\)= \(\frac{3-1}{1.3}\)=\(\frac{3}{1.3}\)- \(\frac{1}{1.3}\)= 1 - \(\frac{1}{3}\)

\(\frac{2}{3.5}\) = \(\frac{5-3}{3.5}\)=\(\frac{5}{3.5}\)- \(\frac{3}{3.5}\)=\(\frac{1}{3}\)- \(\frac{1}{5}\)

\(\frac{2}{5.7}\) = \(\frac{7-5}{5.7}\)= \(\frac{7}{5.7}\)- \(\frac{5}{5.7}\)= \(\frac{1}{5}\)- \(\frac{1}{7}\)

.........

\(\frac{2}{99.101}\) = \(\frac{101-99}{99.101}\)= \(\frac{101}{99.101}\)- \(\frac{99}{99:101}\)= \(\frac{1}{99}\)- \(\frac{1}{101}\)

= 1 - \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{5}\)+ \(\frac{1}{5}\)- \(\frac{1}{7}\)+........ + \(\frac{1}{99}\)- \(\frac{1}{101}\)

= 1- \(\frac{1}{101}\)

= \(\frac{100}{101}\)