Question

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

Answer 1

Ta có:

 2A=1+1/2+1/4+1/8+1/16+1/32+1/64

=> 2A-A=(1+1/2+1/4+1/8+1/16+1/32+1/64) - (1/2+1/4+1/8+1/16+1/32+1/64+1/128)

         A= 1-1/128=127/128

Answer 2

\(\frac{1}{2}+\frac{1}{4}=1-\frac{1}{4}\)

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}=1-\frac{1}{8}\)

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\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}=1-\frac{1}{128}=\frac{127}{128}\)