\(C=\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)
\(2C=\frac{4}{3}+\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}\)
\(2C-C=\frac{4}{3}-\frac{2}{192}\)
\(C=\frac{127}{96}\)
\(C=\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)
\(C=\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\)
\(C=\frac{254}{192}=\frac{127}{96}\)
\(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)
=\(\frac{2}{3}+\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\)
Ta xét phần trong ngoặc :
\(\frac{1}{2}+\frac{1}{4}=1-\frac{1}{4}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}=1-\frac{1}{8}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}=1-\frac{1}{64}\)
=\(\frac{63}{64}\)
=> \(1+\frac{63}{64}=\frac{127}{96}\)
Vậy tổng trên có kết quả là \(\frac{127}{96}\)
k nha
