\(\frac{1}{2}+\frac{1}{6}\)\(+\frac{1}{12}\)\(+...+\frac{1}{9702}\)\(+\frac{1}{9900}\)
= \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}\)\(+...+\frac{1}{98\cdot99}\)+ \(\frac{1}{99\cdot100}\)
= \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\)\(\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
= \(\frac{1}{1}-\frac{1}{100}\)
= \(\frac{100}{100}\)- \(\frac{1}{100}\)
= \(\frac{99}{100}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9702}+\frac{1}{9900}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Gọi dãy trên là A
\(\Leftrightarrow A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\)
\(\Leftrightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Leftrightarrow A=1-\frac{1}{100}+0+...+0\)
\(\Leftrightarrow A=\frac{99}{100}\)
\(S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9702}+\frac{1}{9900}\)
\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(S=1-\frac{1}{100}\)
\(S=\frac{99}{100}\)
1/2+1/6+1/12+.....+1/9072+1/9900
=1/1.2+1/2.3+1/3.4+...+1/98.99+1/99.100
=1/1-1/2+1/2-1/3+1/3-1/4+.....+1/98-1/99+1/99-1/100
=1/1-1/100
=99/100
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9702}+\frac{1}{9900}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{100}=\frac{99}{100}\)