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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)
\(2A=1+\dfrac{1}{2}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}\)
\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{18}+\dfrac{1}{32}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\right)\)
\(A=1-\dfrac{1}{64}\)
\(A=\dfrac{63}{64}\)
giúp em với ạ , em cấn gấp , 4h em học rồi ạ . Cảm ơn những ai giúp em ạ , em tick luôn ạ
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\)
\(A=1-\frac{1}{64}\)
\(A=\frac{63}{64}\)
=32/64+14/64+8/64+4/64+2/64+1/64
=32+14+8+4+2+1/64
=61/64
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}\)
\(=1-\frac{1}{64}\)
\(=\frac{63}{64}\)
![](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
dề có sai ko
B= 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64
B=1/2+1/2^2+1/2^3+1/2^4+1/2^5+1/2^6
=>2B=1+1/2+1/2^2+...1/3^5
=>2B-B=1-1/2^6
=>B=1-1/64
=>B=63/64