Đáp án:
Đặt A=1/3-2/3^2+3/3^3-3/3^4+......+99/3^99-100/3^100
=3A=1-2/3+3/3^2-4/3^3+...+99/3^98-100/3^99
=4A=3A+A=(1-2/3+3/3^2-4/3^3+...+99/3^98-100/3^99)+(1/3-2/3^2+...+99/3^99-100/3^100)
=4A=1-1/3+1/3^2-1/3^3+...+1/3^98-1/3^99-100/3^100
=12A=3-1+...+1/3^97-1/3^98-100/3^99
=16A=4A+12A=(1-1/3+1/3^2-1/3^3+...+1/3^98-1/3^99-100/3^100)+(3-1+...+1/3^97-1/3^98-100/3^99)
=16A=3-101/3^99-100/3^100<3
=>16A<3
=>A<`3/16`
Đề này mình thi vào cuối kì 2 lớp 6
cả trường ko ai làm được
