Question

\(\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+...+\frac{1}{156}+\frac{1}{182}\)

Answer 1

\(A=\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+...+\frac{1}{156}+\frac{1}{182}\) 

\(A=\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+...+\frac{1}{12.13}+\frac{1}{13.14}\)

\(A=\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+...+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}\)

\(A=\frac{1}{8}-\frac{1}{14}\)

\(A=\frac{3}{56}\)

Answer 2

\(\frac{1}{72}+\frac{1}{90}+....+\frac{1}{156}+\frac{1}{182}\)

\(=\frac{1}{8\cdot9}+\frac{1}{9\cdot10}+....+\frac{1}{12\cdot13}+\frac{1}{13\cdot14}\)

\(=\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+...+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}\)

\(=\frac{1}{8}-\frac{1}{14}\)

\(=\frac{3}{56}\)

Answer 3

\(\frac{1}{71}+\frac{1}{90}+\frac{1}{110}+...+\frac{1}{156}+\frac{1}{182}\)

\(=\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+...+\frac{1}{12.13}+\frac{1}{13.14}\)

\(=1-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+....+\frac{1}{13}-\frac{1}{13}-\frac{1}{14}+\frac{1}{14}\)

\(=1-\frac{1}{14}\)

\(=\frac{13}{14}\)

Answer 4

\(\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+.....+\frac{1}{156}+\frac{1}{182}=\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+.....+\frac{1}{12.13}+\frac{1}{13.14}\)

\(=\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+.....+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}\)

\(=\frac{1}{8}-\frac{1}{14}=\frac{14}{112}-\frac{8}{112}=\frac{6}{112}=\frac{3}{56}\)

Answer 5

\(\frac{1}{72}+\frac{1}{90}+...+\frac{1}{156}+\frac{1}{182}=\frac{1}{8\cdot9}+\frac{1}{9\cdot10}+...+\frac{1}{12\cdot13}+\frac{1}{13\cdot14}\)

\(=\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+...+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}\)

\(=\frac{1}{8}-\frac{1}{14}=\frac{7}{56}-\frac{4}{56}=\frac{3}{56}\)