Tính tổng `A= 1+1/2 + (1/2)^2 + (1/2)^3 + … + (1/2)^2022`

2 bình luận về “Tính tổng `A= 1+1/2 + (1/2)^2 + (1/2)^3 + … + (1/2)^2022`”

1. A=1+1/2+(1/2)^2+(1/2)^3+…+(1/2)^2022

   =>1/2 . A=1/2 . [1+1/2+(1/2)^2+(1/2)^3+…+(1/2)^2022]

   1/2A=1/2+(1/2)^2+(1/2)^3+(1/2)^4+…+(1/2)^2023

   Lấy 1/2A-A ta được:

   1/2A-A=[1/2+(1/2)^2+(1/2)^3+(1/2)^4+…+(1/2)^2023]-[1+1/2+(1/2)^2+(1/2)^3+…+(1/2)^2022]

   -1/2 A=(1/2)^2023-1

   A=(-2).[(1/2)^2023-1]

   A=-2. 1/2^2023 +2

   A=(-2)/2^2023+2

   A=(-1)/2^2022+2

   Vậy A=(-1)/2^2022+2

    

   Trả lời
2. Giải đáp + Lời giải và giải thích chi tiết:

   A = 1 + 1/2 + (1/2)^2 + (1/2)^3 + … + (1/2)^2022

   1/2A = 1/2 + (1/2)^2 + (1/2)^3 + (1/2)^4 + … + (1/2)^2023

   A – 1/2A = (1 + 1/2 + (1/2)^2 + (1/2)^3 + … + (1/2)^2022) – (1/2 + (1/2)^2 + (1/2)^3 + (1/2)^4 + … + (1/2)^2023)

   1/2A = 1 + 1/2 + (1/2)^2 + (1/2)^3 + … + (1/2)^2022 -1/2 – (1/2)^2 – (1/2)^3 – (1/2)^4 – … – (1/2)^2023

   1/2A = 1 – (1/2)^2023

   A = (1 – (1/2)^2023) : 1/2

   A = 2 – (1/2)^2022

   Trả lời