Chứng tỏ rằng A<5/4 A=3/1.2.3+5/2.3.4+7/3.4.5+....+201/100.101.102 29/09/2021 Bởi Skylar Chứng tỏ rằng A<5/4 A=3/1.2.3+5/2.3.4+7/3.4.5+....+201/100.101.102
Đáp án: $$A = {{12675} \over {10302}} < {5 \over 4}$$ Giải thích các bước giải: $$\eqalign{ & A = {3 \over {1.2.3}} + {5 \over {2.3.4}} + {7 \over {3.4.5}} + … + {{201} \over {100.101.102}} \cr & A = \left( {{1 \over {1.2.3}} + {1 \over {2.3.4}} + {1 \over {3.4.5}} + … + {1 \over {100.101.102}}} \right) + \left( {{2 \over {1.2.3}} + {4 \over {2.3.4}} + {6 \over {3.4.5}} + … + {{200} \over {100.101.102}}} \right) \cr & A = \left( {{1 \over {1.2.3}} + {1 \over {2.3.4}} + {1 \over {3.4.5}} + … + {1 \over {100.101.102}}} \right) + 2\left( {{1 \over {1.2.3}} + {2 \over {2.3.4}} + {3 \over {3.4.5}} + …{{100} \over {100.101.102}}} \right) \cr & A = B + 2C \cr & B = {1 \over {1.2.3}} + {1 \over {2.3.4}} + {1 \over {3.4.5}} + … + {1 \over {100.101.102}} \cr & 2B = {2 \over {1.2.3}} + {2 \over {2.3.4}} + {2 \over {3.4.5}} + … + {2 \over {100.101.102}} \cr & 2B = \left( {{1 \over {1.2}} – {1 \over {2.3}} + {1 \over {2.3}} – {1 \over {3.4}} + {2 \over {3.4}} – {2 \over {4.5}} + … + {1 \over {100.101}} – {1 \over {101.102}}} \right) \cr & 2B = {1 \over {1.2}} – {1 \over {101.102}} = {{2575} \over {5151}} \cr & B = {{2575} \over {5151}}:2 = {{2575} \over {10302}} \cr & C = {1 \over {1.2.3}} + {2 \over {2.3.4}} + {3 \over {3.4.5}} + …{{100} \over {100.101.102}} \cr & C = {1 \over {2.3}} + {1 \over {3.4}} + {1 \over {4.5}} + … + {1 \over {101.102}} \cr & C = {1 \over 2} – {1 \over 3} + {1 \over 3} – {1 \over 4} + … + {1 \over {101}} – {1 \over {102}} \cr & C = {1 \over 2} – {1 \over {102}} = {{25} \over {51}} \cr & Vay\,\,A = B + 2C = {{2575} \over {10302}} + {{50} \over {51}} = {{2575 + 202.50} \over {5151}} = {{12675} \over {10302}} \cr & {{12675} \over {10302}} = {{12675.2} \over {10302.2}} = {{25350} \over {20604}} \cr & {5 \over 4} = {{5.5151} \over {4.5151}} = {{25755} \over {20604}} \cr & \Rightarrow {{12675} \over {10302}} < {5 \over 4} \Rightarrow A < {5 \over 4}\,\,\left( {dpcm} \right) \cr} $$ Bình luận
Đáp án:
$$A = {{12675} \over {10302}} < {5 \over 4}$$
Giải thích các bước giải:
$$\eqalign{
& A = {3 \over {1.2.3}} + {5 \over {2.3.4}} + {7 \over {3.4.5}} + … + {{201} \over {100.101.102}} \cr
& A = \left( {{1 \over {1.2.3}} + {1 \over {2.3.4}} + {1 \over {3.4.5}} + … + {1 \over {100.101.102}}} \right) + \left( {{2 \over {1.2.3}} + {4 \over {2.3.4}} + {6 \over {3.4.5}} + … + {{200} \over {100.101.102}}} \right) \cr
& A = \left( {{1 \over {1.2.3}} + {1 \over {2.3.4}} + {1 \over {3.4.5}} + … + {1 \over {100.101.102}}} \right) + 2\left( {{1 \over {1.2.3}} + {2 \over {2.3.4}} + {3 \over {3.4.5}} + …{{100} \over {100.101.102}}} \right) \cr
& A = B + 2C \cr
& B = {1 \over {1.2.3}} + {1 \over {2.3.4}} + {1 \over {3.4.5}} + … + {1 \over {100.101.102}} \cr
& 2B = {2 \over {1.2.3}} + {2 \over {2.3.4}} + {2 \over {3.4.5}} + … + {2 \over {100.101.102}} \cr
& 2B = \left( {{1 \over {1.2}} – {1 \over {2.3}} + {1 \over {2.3}} – {1 \over {3.4}} + {2 \over {3.4}} – {2 \over {4.5}} + … + {1 \over {100.101}} – {1 \over {101.102}}} \right) \cr
& 2B = {1 \over {1.2}} – {1 \over {101.102}} = {{2575} \over {5151}} \cr
& B = {{2575} \over {5151}}:2 = {{2575} \over {10302}} \cr
& C = {1 \over {1.2.3}} + {2 \over {2.3.4}} + {3 \over {3.4.5}} + …{{100} \over {100.101.102}} \cr
& C = {1 \over {2.3}} + {1 \over {3.4}} + {1 \over {4.5}} + … + {1 \over {101.102}} \cr
& C = {1 \over 2} – {1 \over 3} + {1 \over 3} – {1 \over 4} + … + {1 \over {101}} – {1 \over {102}} \cr
& C = {1 \over 2} – {1 \over {102}} = {{25} \over {51}} \cr
& Vay\,\,A = B + 2C = {{2575} \over {10302}} + {{50} \over {51}} = {{2575 + 202.50} \over {5151}} = {{12675} \over {10302}} \cr
& {{12675} \over {10302}} = {{12675.2} \over {10302.2}} = {{25350} \over {20604}} \cr
& {5 \over 4} = {{5.5151} \over {4.5151}} = {{25755} \over {20604}} \cr
& \Rightarrow {{12675} \over {10302}} < {5 \over 4} \Rightarrow A < {5 \over 4}\,\,\left( {dpcm} \right) \cr} $$