Tính :
a) \(A = {1 \over {1.2}} + {1 \over {2.3}} + {1 \over {3.4}} + … + {1 \over {98.99}} + {1 \over {99.100}}\)
b) \(B = {3 \over {2.5}} + {3 \over {5.8}} + … + {3 \over {17.20}}\)
\(\eqalign{ & A = {1 \over {1.2}} + {1 \over {2.3}} + {1 \over {3.4}} + … + {1 \over {98.99}} + {1 \over {99.100}} \cr & = {{2 – 1} \over {1.2}} + {{3 – 2} \over {2.3}} + {{4 – 3} \over {3.4}} + … + {{99 – 98} \over {98.99}} + {{100 – 99} \over {99.100}} \cr & = 1 – {1 \over 2} + {1 \over 2} – {1 \over 3} + {1 \over 3} – {1 \over 4} + …. + {1 \over {98}} – {1 \over {99}} + {1 \over {99}} – {1 \over {100}} \cr & = 1 – {1 \over {100}} = {{100} \over {100}} – {1 \over {100}} = {{99} \over {100}}. \cr & B = {3 \over {2.5}} + {3 \over {5.8}} + … + {3 \over {17.20}} \cr & = {{5 – 2} \over {2.5}} + {{8 – 5} \over {5.8}} + … + {{20 – 17} \over {17.20}} \cr & = {1 \over 2} – {1 \over 5} + {1 \over 5} – {1 \over 8} + … + {1 \over {17}} – {1 \over {20}} \cr & = {1 \over 2} – {1 \over {20}} = {{10} \over {20}} – {1 \over {20}} = {9 \over {20}}. \cr} \)