Tính tổng A = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64
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![](https://rs.olm.vn/images/avt/0.png?1311)
nhân A lên 2A sau đó lấy 2A-A là đc :
2A =1+1/2+.....+1/32
2A-A=(1+1/2+.....+1/32)-(1/2+1/4+.....+1/32+1/64)
A=1-1/64
A=63/64
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(2A=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\times2\)
\(2A=\frac{1}{2}\times2+\frac{1}{4}\times2+\frac{1}{8}\times2+\frac{1}{16}\times2+\frac{1}{32}\times2+\frac{1}{64}\times2\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\)
\(A=1-\frac{1}{64}\)
\(A=\frac{63}{64}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}$
$2\times A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}$
$2\times A-A=(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32})-(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64})$
$A=1-\frac{1}{64}=\frac{63}{64}$
![](https://rs.olm.vn/images/avt/0.png?1311)
Cộng thêm 1/2 vào biểu thức đã cho, có:
S + 1/2= 1/2+1/4+ 1/8+ 1/16+1/32+1/64+1/128
Nhận xét:
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : A = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64
=> 2A = 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32
=> 2A - A = 1 - 1/64
=> A = 1 - 1/64
=> A = 63/64
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(\Rightarrow2A-A=1-\frac{1}{128}=\frac{127}{128}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
- Tính A x 2
A x 2 = 1 + 1/2 +1/4 +1/8 +1/16+ 1/32
- Tính A bằng cách A = A x 2 – A
Vậy A = 1 + 1/2 +1/4 +1/8 +1/16+ 1/32 - 1/2 - 1/4 - 1/8 -1/16 - 1/32 - 1/64
A = 1 - 1/64
A = 63/64
Đặt A=1/2+1/8+1/16+1/32+1/64.Ta có:
A x 2=(1/2+1/8+1/16+1/32+1/64) x 2
A x 2=1+1/2+1/4+1/8+1/16+1/32
A x 2-A=(1+1/2+1/4+1/8+1/16+1/32)-(1/2+1/4+1/8+1/16+1/32+1/64)
A=1-1/64=63/64
Ai đi qua cho mình xin cái k mình k lại cho
![](https://rs.olm.vn/images/avt/0.png?1311)
S=1/2+1/4+1/8+1/16+1/32+1/64
S=1-1/2+1/2-1/4+1/4-1/8+1/8-1/16+1/16-1/32+1/32-1/64
S=1-1/64
S=63/64
![](https://rs.olm.vn/images/avt/0.png?1311)
Tính không quy đồng mẫu:
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{32}-\frac{1}{64}\)
\(A=1-\frac{1}{64}=\frac{63}{64}\)
Ta có : \(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{64}\)
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^6}\)
\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^5}\)
\(\Rightarrow2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^5}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^6}\right)\)
\(\Rightarrow A=1-\dfrac{1}{2^6}=1-\dfrac{1}{64}=\dfrac{63}{64}\)
\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}+...+\dfrac{1}{32}-\dfrac{1}{64}\)
\(=1-\dfrac{1}{64}\)
\(=\dfrac{63}{64}\)