X x (1/2+1/4+1/8+1/16+1/32+1/64+1/128) = 127/128

X x 127/128                                            = 127/128

                                                        X   = 127/128 : 127/128

                                                        X   = 1

b: A=1/3+1/9+...+1/3^10

=>3A=1+1/3+...+1/3^9

=>A*2=1-1/3^10=(3^10-1)/3^10

=>A=(3^10-1)/(2*3^10)

c: C=3/2+3/8+3/32+3/128+3/512

=>4C=6+3/2+...+3/128

=>3C=6-3/512

=>C=1023/512

d: A=1/2+...+1/256

=>2A=1+1/2+...+1/128

=>A=1-1/256=255/256

chào bn

Help

2

4

8

16

32

64

128

256

2

4

8

16

64

128

256

          k mk nhé

1+1=2

2+2=4

4+4=8

8+8=16

16+16=32

32+32=64

64+64=128

128+128=256

tk m nhé

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

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{4}+........+\frac{1}{64}\)

Lấy 2A-A ta có:

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

\(\Rightarrow A=1-\frac{1}{128}\)

\(\Rightarrow A=\frac{127}{128}\)

Bạn học lớp 5 mấy vậy? Trường nào vậy?

1/2+1/4+1/8+1/16+....+1/128

=1-1/2+1/2-1/4+1/4-1/8+1/8-1/16+...+1/64-1/128

=1-1/128

=127/128

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

=127/128

= 1 - 1/2 + 1/2 - 1/4 + 1/4 - 1/8 +...+1/64 - 1/128                                                                                                                                 = 1 - 1/128                                                                                                                                                                                       = 127/128

1+1=2

2+2=4

4+4=8

8+8=16

16+16=32

32+32=64

64+64=128

128+128=256

2

4

8

16

32

64

128

256

1+1=2

2+2=44+4=8

8+8=16

16+16=32

32+32=64

64+64=128

128+128=256

C= \(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\)

2C = \(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)

2C-C = \(1-\dfrac{1}{128}\)

C= \(\dfrac{127}{128}\)

C= 12+14+18+116+132+164+112821​+41​+81​+161​+321​+641​+1281​

2C = 1+12+14+18+116+132+1641+21​+41​+81​+161​+321​+641​

2C-C = 1−11281−1281​

C= 127128128127​
 

Đặt: \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}+\frac{1}{256}\)

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

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\)

\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^2}\right)-\left(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\right)\)

\(\Rightarrow A=1-\frac{1}{2^7}\)

E= 1/2 + 1/4 + 1/8 + 1/16 + ... + 1/128 + 1/256

2E = 2 ( 1/2 + 1/4 + 1/8 + 1/16 + ... + 1/128 + 1/256 )

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

=> E = 2E - E

= (1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128) - (1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 )

= 1 - 1/256

= 255/256

k nhá, thanks

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

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

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

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

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

\(E=1-\frac{1}{256}\)

\(E=\frac{255}{256}\)