Question

[Junjou Vuong]

1/2+1/4+1/8+1/16+1/32+1/64+1/128

Answer 1

[Vanlacongchua]

Gọi biểu thức trên là A ta có

\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)

=> \(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)

=> \(A=2A-A\)\(=1-\frac{1}{128}\)

Vậy \(A=1-\frac{1}{128}\)