K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

a.1025

mk chịu câu b

*** mk nha

3 tháng 7 2019

1) 1/1.2 + 1/2.3 + ... + 1/6.7

= 1 - 1/2 + 1/2 - 1/3 + ... + 1/6 - 1/7

= 1 - 1/7

= 6/7

2) 1/2 + 1/6 + 1/12 + .. + 1/72

= 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/8.9

= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/8 - 1/9

= 1 - 1/9

= 8/9

3) \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2019}\right)\)

\(\frac{1}{2}.\frac{2}{3}...\frac{2019}{2020}\)

\(\frac{1.2....2019}{2.3...2020}\)

\(\frac{1}{2020}\)

4) A = \(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{512}\)

       = \(\frac{1}{2^2}+\frac{2}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^9}\)

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

Lấy 2A - A = \(\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^9}\right)\)

             A  = \(\frac{1}{2}-\frac{1}{2^9}\)

8 tháng 8 2019

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)

\(< =>\frac{128}{256}+\frac{64}{256}+\frac{32}{256}+\frac{16}{256}+\frac{8}{256}+\frac{4}{256}+\frac{2}{256}+\frac{1}{256}\)

\(< =>\frac{128+64+32+16+8+4+2+1}{256}\)

\(< =>\frac{255}{256}\)

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

\(< =>\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(< =>\frac{1}{1}-\frac{1}{100}\)

\(< =>\frac{99}{100}\)

\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{100}\right)\)

\(< =>\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{99}{100}\)

\(< =>\frac{1\cdot2\cdot3\cdot...\cdot99}{2\cdot3\cdot4\cdot...\cdot100}\)

\(< =>\frac{1}{100}\)

mk chuc ban hoc tot nhe :))

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/ 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

3 tháng 8 2023

Đặt A = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256

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

=> 2A - A = (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)

=> A = 1 - 1/256

=> A = 255/256

Vậy: ...

7 tháng 8 2016

các cậu trả lời giúp mình đi

1 + 1 = 2

2 + 2 = 4

4 + 4 = 8

8 + 8 = 16

16 + 16 = 32

32 + 32 = 64

64 + 64 = 128

128 + 128 = 256

256 + 256 = 512

512 + 512 = 1024

1024 + 1024 = 2048

16 tháng 8 2020

1+1=2

2+2=4

4+4=8

8+8=16

16+16=32

32+32=64

64+64=128

128+128=265

256+256=512

512+512=1024

1024+1024=2048

Học tốt ^_^

18 tháng 4 2016

1+1=2

2+2=4

4+4=8

8+8=16

16+16=32

32+32=64

64+64=128

128+128=256

256+256=512

512+512=1024

18 tháng 4 2016

1+1=2

2+2=4

4+4=8

8+8=16

16+16=32

32+32=64

64+64=128

128+128=156

256+256=512

512+512=1024

2 tháng 4 2020

\(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+...+\frac{1}{512}-\frac{1}{1024}\)

\(A=\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+...+\frac{1}{2^9}-\frac{1}{2^{10}}\)

\(2A=1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+...+\frac{1}{2^8}-\frac{1}{2^9}\)

\(3A=1-\frac{1}{2^{10}}< 1\)

\(\Rightarrow A< \frac{1}{3}\)