Question

B=\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+10}\)

C=\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)

 

Answer 1

Ta có công thức : \(1+2+3+...+n=\frac{n.\left(n+1\right)}{2}\)

\(\Rightarrow B=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+10}\)

           \(=\frac{1}{\frac{\left(1+2\right).2}{2}}+\frac{1}{\frac{\left(1+3\right).3}{2}}+...+\frac{1}{\frac{\left(1+10\right)10}{2}}\)

           \(=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{10.11}\)

           \(=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)

           \(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)

           \(=2.\left(\frac{1}{2}-\frac{1}{11}\right)=2.\frac{9}{22}=\frac{9}{11}\)

Answer 2

B=9/11

Answer 3

B=9/11

Answer 4

C = \(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}\)\(+\frac{89}{90}\)

\(C=\frac{1\cdot2-1}{1\cdot2}+\frac{2\cdot3-1}{2\cdot3}+\frac{3\cdot4-1}{3\cdot4}+...+\frac{9\cdot10-1}{9\cdot10}\)

\(C=1-\frac{1}{1\cdot2}+1-\frac{1}{2\cdot3}+1-\frac{1}{3\cdot4}+...+1+\frac{1}{9\cdot10}\)

\(C=9-\left(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{9\cdot10}\right)\)

\(C=9-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(C=9-\left(1-\frac{1}{10}\right)=9-\frac{9}{10}=\frac{81}{10}\)