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

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

= 1 – 1/256

= 255/256

=> a/256 = 255/256

=> a = 255.

Vậy a = 255

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

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: ...

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

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

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

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

\(A=\frac{511}{512}\)
 

      A =          \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{64}\) + \(\dfrac{1}{128}\) + \(\dfrac{1}{256}\)

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

2A - A =   1 - \(\dfrac{1}{256}\)

       A   = \(\dfrac{255}{256}\)

a.\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2015}\right)=\frac{1}{2}.\frac{2}{3}...\frac{2014}{2015}=\frac{1.2.3...2014}{2.3...2015}=\frac{1}{2015}\)

b.\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{128}-\frac{1}{256}=1-\frac{1}{256}=\frac{255}{256}\)

c.\(\frac{5}{2}+\frac{5}{4}+\frac{5}{8}+...+\frac{5}{256}=5\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\right)=5.\frac{255}{256}=\frac{1275}{256}\)

d.14,35+(13,7-13,6).1=14,35+0,1.1=14,35+0,1=14,45

a.1/2015

a) \(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{256}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}+...-\dfrac{1}{128}+\dfrac{1}{128}-\dfrac{1}{256}\)

\(=1-\dfrac{1}{256}\)

\(=\dfrac{255}{256}\)

b) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{13.14}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{14}\)

\(=1-\dfrac{1}{14}\)

\(=\dfrac{13}{14}\)

c) \(\dfrac{3}{15.18}+\dfrac{3}{18.21}+\dfrac{3}{21.24}+...+\dfrac{3}{87.90}\)

\(=3.\left(\dfrac{1}{15.18}+\dfrac{1}{18.21}+\dfrac{1}{21.24}+...+\dfrac{1}{87.90}\right)\)

\(=3.\left[\dfrac{1}{3}.\left(\dfrac{1}{15}-\dfrac{1}{18}\right)+\dfrac{1}{3}.\left(\dfrac{1}{18}-\dfrac{1}{21}\right)+\dfrac{1}{3}.\left(\dfrac{1}{21}-\dfrac{1}{24}\right)+...+\dfrac{1}{3}.\left(\dfrac{1}{87}-\dfrac{1}{90}\right)\right]\)

\(=3.\dfrac{1}{3}.\left(\dfrac{1}{15}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{24}+...+\dfrac{1}{87}-\dfrac{1}{90}\right)\)

\(=\dfrac{1}{15}-\dfrac{1}{90}\)

\(=\dfrac{6}{90}-\dfrac{1}{90}\)

\(=\dfrac{5}{90}=\dfrac{1}{18}\)

tớ đang cần gấp

a)255/256
b)13/14
c) 1/18

 Ta có: A =1/2+1/4+1/8+1/16+....+1/256+1/512 

=> 2A = 1 + 1/2 + 1/4 + 1/8 + ...+ 1/128 + 1/256

=> 2A - A = (1 + 1/2 + 1/4 + 1/8 + ...+ 1/128 + 1/256 -(1/2+1/4+1/8+1/16+....+1/256+1/512 )

         A =  1 - 1/512 = 511/512