Cho A=1/2+1/4+1/8+1/16+.....+1/256.Tìm A
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
vì quá dễ nên mình không thể trả lời bạn được nhé!
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 1/2 + 1/4 + 1/8 + ... + 1/1024
2A = 1 + 1/2 + 1/4 + ... + 1/512
2A - A = (1 + 1/2 + 1/4 + ... + 1/512) - (1/2 + 1/4 + 1/8 + ... + 1/1024)
A = 1 - 1/1024
A = 1023/1024
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{1024}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{4}+......+\frac{1}{512}\)
\(\Rightarrow A=2A-A=1-\frac{1}{1024}\)
\(A=\frac{1023}{1024}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A = \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{64}\) + \(\dfrac{1}{128}\) + \(\dfrac{1}{256}\)
2A = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{64}\) + \(\dfrac{1}{128}\)
2A - A = 1 - \(\dfrac{1}{256}\)
A = \(\dfrac{255}{256}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn kiểm tra lại đề hộ. Nếu có phân số \(\frac{1}{4}\)thì chịu còn không có thì dễ.
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: A =1/2+1/4+1/8+1/16+....+1/256+1/512
=> 2A = 1 + 1/2 + 1/4 + 1/8 + ...+ 1/128 + 1/256
=> 2A - A = (1 + 1/2 + 1/4 + 1/8 + ...+ 1/128 + 1/256 -(1/2+1/4+1/8+1/16+....+1/256+1/512 )
A = 1 - 1/512 = 511/512
![](https://rs.olm.vn/images/avt/0.png?1311)
a: 4A=4+4^2+...+4^9
=>3A=4^9-1
=>A=(4^9-1)/3
b: 2A=1+1/2+...+1/2^7
=>A=1-1/256=255/256
c: =1-1/5+1/5-1/9+...+1/85-1/89
=1-1/89=88/89
d: =1/3(3/1*4+3/4*7+...+3/304*307)
=1/3(1-1/4+1/4-1/7+...+1/304-1/307)
=1/3*306/307=102/307
e: E=1-1/2+1/2-1/3+...+1/11-1/12
=1-1/12=11/12
g: =2/5(1-1/6+1/6-1/11+...+1/96-1/101)
=2/5*100/101=40/101
![](https://rs.olm.vn/images/avt/0.png?1311)
đề phải là 1 +1/2 + 1/4 +1/32 + 1/64 + 1/128 +1/256 +/512 +1/1024 moi dug