M=(1/5+1/5^2+1/5^3+...+1/5^2023) + 1/5x(1/5+1/5^2+1/5^3+...+1/5^2022) + ... + 1/5^2021x(1/5+1/5^2) + 1/5^2022x1/5
Xét biểu thức N=1/5+1/5^2+1/5^3 + ... + 1/5^k (K>0, k thuộc Z)
=> 5N=1+1/5+1/5^2+1/5^3+...+1/5^(k-1)
=> 4N= 5N - N =1 - 1/5^k
=> 1/5+1/5^2+1/5^3 + ... + 1/5^k = 1/4x(1-1/5^k)
Thay vào biểu thức M, ta có:
M= 1/4x(1-1/5^2023) + 1/5x1/4x(1-1/5^2022) + ... + 1/5^2021x1/4x(1-1/5^2) + 1/5^2022x1/4x(1-1/5)
=> 4M = (1+1/5+1/5^2+...+1/5^2022) - 2023/5^2023
=> 4M = 5/4x(1-1/5^2023)-2023/5^2023 < 5/4
=> M < 5/16 < 1/3
Vậy M < 1/3 [ vượt chỉ tiêu nhé =)) ]