Câu hỏi của Sir Nghi

Question

1. So sánh a) $A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}$ và B= $\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{13}{60}$b) \(C=\dfrac{2019}{2021}+\dfr...

bui duy phu · Accepted Answer

a) Ta có:2A=2.(12+122+123+...+122020+122021)2�=2.12+122+123+...+122  020+122  0212A=1+12+122+123+...+122019+1220202�=1+12+122+123+...+122  019+122  020Suy ra: 2A−A=(1+12+122+123+...+122019+122020)2�−�=1+12+122+123+...+122  019+122  020                             −(12+122+123+...+122020+122021)−12+122+123+...+122  020+122  021Do đó A=1−122021<1�=1−122021<1.Lại có B=13+14+15+1360=20+15+12+1360=6060=1�=13+14+15+1360=20+15+12+1360=6060=1.Vậy A < B. Câu trả lời: a) Ta có:2A=2.(12+122+123+...+122020+122021)2�=2.12+122+123+...+122  020+122  0212A=1+12+122+123+...+122019+1220202�=1+12+122+123+...+122  019+122  020Suy ra: 2A−A=(1+12+122+123+...+122019+122020)2�−�=1+12+122+123+...+122  019+122  020                             −(12+122+123+...+122020+122021)−12+122+123+...+122  020+122  021Do đó A=1−122021<1�=1−122021<1.Lại có B=13+14+15+1360=20+15+12+1360=6060=1�=13+14+15+1360=20+15+12+1360=6060=1.Vậy A < B.

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