\(A=-\frac{1}{20}+-\frac{1}{30}+-\frac{1}{42}+-\frac{1}{56}+-\frac{1}{72}+-\frac{1}{90}\)

\(\Rightarrow A=-1\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{9.10}\right)\)

\(A=-1\left(\frac{1}{4}-\frac{1}{10}\right)\)

\(\Rightarrow A=-\frac{3}{20}\)

\(A=\frac{-1}{20}-\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)

\(A=-\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)

\(A=-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(A=-\left(\frac{1}{4}-\frac{1}{10}\right)\)

\(A=\frac{-3}{20}\)

#

Cách 1:

B=1/2+1/4+1/8+1/16+1/32+1/64

B=1-1/2 + 1/2-1/4 + 1/4-1/8 +1/8-1/16 + 1/16-1/32 + 1/32-1/64

B=1-1/64

B=63/64

Cách 2:

B=1/2+1/4+1/8+1/16+1/32+1/64

B=1/2¹+1/2²+1/2³+1/2⁴+1/2⁵+1/2⁶

2B=1+1/2¹+1/2^2+1/2^3+1/2^4+1/2^5

2B-B=1-1/2^6

B=1-1/64 

B=63/64

Đặt 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 + 1/2 + 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

Theo đề bài ta có :

\(2B=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{128}\)

\(\Leftrightarrow2B-B=\left(1+\frac{1}{2}+...+\frac{1}{128}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\right)\)

\(\Leftrightarrow B=1-\frac{1}{256}\)

\(\Leftrightarrow B=\frac{255}{256}\)

\(B=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+..+\frac{1}{256}\)

\(\Rightarrow B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+..+\frac{1}{2^8}\)

\(\Rightarrow2B=1+\frac{1}{2}+\frac{1}{2^2}+..+\frac{1}{2^7}\)

\(\Rightarrow2B-B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\right)\)

\(\Rightarrow B=1-\frac{1}{2^8}\)

Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)

\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)\)

\(A=1-\frac{1}{32}=\frac{31}{32}\)

31/32 nhé bạn

=\(\frac{31}{32}\)

Đặt biểu thức trên là A

Ta có:

 \(2A-A=A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)=1-\frac{1}{32}=\frac{31}{32}\)

Cách 1:

Đặt A = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\)

2A = \(1+\frac{1}{2}+\frac{1}{4}+....+\frac{1}{64}\)

A = 2A - A = \(1-\frac{1}{128}\)

=> A = \(\frac{127}{128}\)

Cách 2:

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\)

= \(\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+...+\left(\frac{1}{64}-\frac{1}{128}\right)\)

= \(1-\frac{1}{128}\)

= \(\frac{127}{128}\)

1/2 - 1/4 + 1/4 - 1/8 + 1/8 - 1/16 + 1/16 - 1/32 + 1/32 - 1/64 + 1/64 - 1/128

Gạch 1/4 với 1/4 , 1/8 với 1/8 , 1/16 với 1/16 , 1/32 với 1/32 , 1/64 với 1/64

Còn 1/2 - 1/128 = 63/128 

Đúng thì k cho mình 

  \(\frac{1}{2}\)+ \(\frac{1}{4}\) + \(\frac{1}{8}\) + \(\frac{1}{16}\) + \(\frac{1}{32}\)

= [ 1 - \(\frac{1}{2}\)] + [ \(\frac{1}{2}\) - \(\frac{1}{4}\)] + [ \(\frac{1}{4}\) - \(\frac{1}{8}\)] + [ \(\frac{1}{8}\) - \(\frac{1}{16}\)] + [ \(\frac{1}{16}\) - \(\frac{1}{32}\)]

Xóa bỏ các phân số trùng lặp , ta được tổng của dãy số là :

1 - \(\frac{1}{32}\) = \(\frac{31}{32}\)

   Đ/S :\(\frac{31}{32}\)

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 + ... + 1/32 = 1 - 1/32 = 31/32

Vậy 1/2 + 1/4 + 1/8  +1/16 + 1/32 = 31/32

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 = \(1-\frac{1}{64}\)

=> A = \(\frac{63}{64}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{32}-\frac{1}{64}\)

\(=1-\frac{1}{64}\)

=\(=\frac{63}{64}\)