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