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\(A=\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-\frac{1}{32}-\frac{1}{64}\)
\(\Leftrightarrow\)\(2A=1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-\frac{1}{32}\)
\(\Leftrightarrow\)\(2A-A=\left(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-\frac{1}{32}\right)\)\(-\left(\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-\frac{1}{8}-\frac{1}{16}-\frac{1}{32}-\frac{1}{64}\right)\)
\(\Leftrightarrow\)\(A=1-\frac{1}{64}\)\(=\frac{63}{64}\)
1/2 - 1/4 - 1/8 - 1/16 - 1/32 - 1/64
= 32/64 - 16/64 - 8/64 - 4/64 - 2/64 - 1/64
= 1/64 .
^ - ^ . Mình không chăc chắn lắm đâu !
Tính không quy đồng mẫu:
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{32}-\frac{1}{64}\)
\(A=1-\frac{1}{64}=\frac{63}{64}\)
A = 1 - 1/2 + 1/2 - 1/4 + 1/4 - 1/8 + 1/8 - 1/16 + 1/16 - 1/32 + 1/32 - 1/64
A = 1 - 1/64
A = 63/64
A = 1/2 + 1/4 +1/8+ 1/16 +1/32 +1/64
A = 1- 1/2 + 1/2 - 1/4 + 1/4 - 1/8 + 1/8 - 1/16 + 1/16 - 1/32 + 1/32 - 1/64
A = 1 - 1/64
A = 63/64
\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)
\(\dfrac{4}{2}A=\dfrac{4}{2}\cdot\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\right)\)
\(2A=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}\)
\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\right)\)
\(A=\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+..\left(\dfrac{1}{32}-\dfrac{1}{32}\right)+\left(1-\dfrac{1}{64}\right)\)
\(A=1-\dfrac{1}{64}\)
\(A=\dfrac{63}{64}\)
ta thấy:
1/2 + 1/4 = 3/4 = 1 - 1/4
1/2 + 1/4 + 1/8 = 7/8 = 1 - 1/8
........................
=> 1/2 + 1/4 +...+ 1/134 = 1 - 1/134 = 133/134
kq = \(\frac{127}{128}\)Bạn chỉ cần bấm máy tính là ra bài này dễ mà hihi :D :))
63/64 nhé
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{32}+\frac{1}{64}\)
=\(\frac{1}{1\cdot2}+\frac{1}{2\cdot2}+\frac{1}{2\cdot4}+\frac{1}{2\cdot16}+\frac{1}{2\cdot32}\)
=\(\frac{1}{2}\cdot\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{16}+\frac{1}{32}\right)\)
=\(\frac{1}{2}\cdot\frac{119}{64}\)
=\(\frac{119}{128}\)