Vậy dãy số đó là:

1/2 + 1/4 + 1/8 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512 =

256/512 + 128/512 + 64/512 + 32/512 + 16/512 + 8/512 + 4/512 + 2/512 + 1/512 511/512

Đáp số: 511/512

1/2 + 1/4 =3/4

1/2+1/4 +1/8=7/8

1/2+1/4+1/8+1/16=15/16

....tương tự ta có

1/2 +1/4+1/8+1/16+...+1/256+1/512=511/512

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{256}+\frac{1}{512}\)

\(=\frac{256}{512}+\frac{128}{512}+\frac{64}{512}+\frac{32}{512}+\frac{16}{512}+\frac{8}{512}+\frac{4}{512}+\frac{2}{512}\)

\(=\frac{511}{512}\)

Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{256}+\frac{1}{512}\)

\(Ax2=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{256}\)

\(Ax2-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{256}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{256}+\frac{1}{512}\right)\)

\(A=1-\frac{1}{512}\)

\(A=\frac{511}{512}\)
 

2A=1+1/2+1/4+1/8+.........+1/512

2A-A=(1+1/2+1/4+1/8+....+1/512)-(1/2+1/4+1/8+.....+1/1024)

A=1-1/1024

A=1023/1024

vậy A=1023/1024

A = (1/2+1/64)+(1/4+1/128)+(1/8+1/256)+(1/16+1/512)+(1/32+1/1024)
= 1/2(1+1/32)+1/4(1+1/32)+1/8(1+1/32)+1/16(1+1/32)+1/32(1+1/32)
= (1+1/32) x (1/2+/14+1/8+1/16+1/32)
= (33/32) x (31/32) = 1023/1024

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

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

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

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

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

\(\Rightarrow A=\frac{1023}{1024}\)

1 + 2 + 3 + 4 + ... + 2108

= (2108 + 1).2018 : 2

= 2019.1009

= 2037171

1 + 4 + 7 + ... + 100

số số hạng là :

(100 - 1) : 3 + 1  = 34

tổng :

1 + 4 + 7 + ... + 100

= (100 + 1).34 : 2

= 101.17

= 1717

đặt A = 1 + 2 + 4 + 8 + 16 + ... + 512

A = 1 + 2 + 2² + 2³ + 2⁴ + ... + 2⁹

2A = 2 + 2² + 2³ + ... + 2¹⁰

2A - A = (2 + 2² + 2³ + ... + 2¹⁰) - (1 + 2 + 2² + ... + 2⁹)

A = 2¹⁰ - 1

bằng 511/512

Dãy số đó có số số hạng là :

( 1/1024 - 1 ) : 

( 1 + 1/1024 ) * 

  1 + 2 + 3 + 4 + 2017 + 2018

2S = 2019 + 2019 + 2019 + ... + 2019(có  số hạng)

  S = 2019 x 2018 : 2 

  S = 2037881

  1 + 4 + 7 + ...+ 100

2S= 101 + 101 +...+101(có 34 số hạng)

  S= 101 x 34 : 2 = 1717

minh cho cong thuc ban tu giai nha 

[(so dau + so cuoi) x so so hang ]/2

trả lời chính xác đi mà

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

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

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}\right)\)

\(A=1-\frac{1}{512}=\frac{511}{512}\)

Đặt:  \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)

\(\Rightarrow2A=2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\right)\)

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)

\(\Rightarrow2A-A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)\(-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\right)\)

\(\Rightarrow A=1-\frac{1}{512}\)

\(\Rightarrow A=\frac{511}{512}\)

~ rất vui vì giúp đc bn ~

ta gọi tổng trên là H

  Ta có:   H=1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256

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

⇒2H−H=(1+1/2+1/4+.........+1/128)−(1/2+1/4+1/8+...........+1/256)

=>H= 1 + 1/2+1/4+1/8+1/16+1/32+1/64-1/2-1/4-/18-1/16-1/32-1/64-1/128-1/256

⇒H=1−1/256

⇒H=255/256

1/2 + 1/4 + 1/8 + 1/16 + ... + 1/256 + 1/512

= 256/512 + 128/512 + 64/512 + ... + 2/512 + 1/512

= 256 + 128 + 64 + .. + 2 + 1 / 512

= ???????

s=1/2+1/4+1/8+1/16+.....+1/256+1/512

sx2=(1/2+1/4+1/8+1/16+....+1/256+1/512)x2

sx2=1+1/2+1/4+1/8+......+1/126+1/256

sx2-s=(1+1/2+1/4+1/8+......+1/256)-(1/2+1/4+1/8+1/16++.....+1/256+1/512)

1+1/2+1/4+1/8+......+1/256-1/2-1/4-1/8-1/16-.....1/256-1/512

=1-1/512=511/512

có ai bik B=1/4+1/8+1/16+1/...+1/256 ko (nếu bik thì giải tử tế nhé mik cảm ơn)