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

1+5/4+5/8+5/16+5/32+5/64=

Trả lời cái kiểu gì thế 

lớp 4 mà khó như lớp 5 phét vừa thôi ông nha

\(1+\frac{5}{4}+\frac{5}{8}+\frac{5}{16}+\frac{5}{32}+\frac{5}{64}\)

\(=\frac{64}{64}+\frac{80}{64}+\frac{40}{64}+\frac{20}{64}+\frac{10}{64}+\frac{5}{64}\)

\(=\frac{64+80+40+20+10+5}{64}\)

\(=\frac{219}{64}\)

\(=\frac{27}{8}\)

1+5/4+5/8+5/16+5/32+5/64

=1+5/4+5/8+5/16+5/32+5/64

=1+(5/4+5/8+5/16+5/32+5/64)

=1+[5x(1/4+1/8+1/16+1/32+1/64)]

A=1/4+1/8+1/16+1/32+1/64

2A=1/2+1/4+1/8+1/16+1/32

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

A=1/2-1/64

A=31/64

1+[5x31/64]

=1+155/64

=219/64

1/ 2 + 2 = 4

2/ 4 + 4 = 8

3/ 8 + 8 = 16

4/ 16 + 16 = 32

5/ 32 + 32 =64

6/ 64 + 64 =128

7/ 128 + 128 =256

8/  256 + 256 =512

9/ 521 + 512 =1033

10/ 2048 + 2048 =4096

k mình đi

4

8

16

32

64

128

256

512

2048

4096

Đặt \(A=1+\frac{5}{4}+\frac{5}{8}+\frac{5}{16}+\frac{5}{32}+\frac{5}{64}\)

\(=5\cdot\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\)

Đặt \(B=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)

\(\Rightarrow2\cdot B=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)

\(\Rightarrow B=2\cdot B-B=1-\frac{1}{64}=\frac{63}{64}\)

\(\Rightarrow A=5\cdot\frac{63}{64}=\frac{315}{64}\)

1+80/64+40/60+20/64+10/64+5/64

1+2 27/64

3 27/64

= 2,421875 bạn nhé

các bạn cho mình cách giải cụ thể đc ko

219/64 nhé