Giải
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\)\(\frac{1}{96}\)
\(=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{12}\)\(+\frac{1}{12}-\frac{1}{24}+\frac{1}{24}-\frac{1}{48}\)\(+\frac{1}{48}-\frac{1}{96}\)
\(=\frac{1}{3}-\frac{1}{96}\)
\(=\frac{31}{96}\)
1/6 + 1/12 + 1/24 + 1/48 + 1/96
= 1/3 - 1/6 + 1/6 - 1/12 + 1/12 - 1/24 + 1/24 - 1/48 + 1/48 - 1/96
= 1/3 - 1/96
= 31/96
Trả lời:
1/6 + 1/12 + 1/24 + 1/48 + 1/96
= 1/3 - 1/6 + 1/6 - 1/12 + 1/12 - 1/24+ 1/24 - 1/48 + 1/48 - 1/96
= 1/3 - 1/96
= 31/96
#hien#