\(A\cdot2=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...+\frac{1}{256}\right)\cdot2\)

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

\(A\cdot2-A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...+\frac{1}{128}-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\right)\)

\(A=1-\frac{1}{256}=\frac{255}{256}\)

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

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

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

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

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

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

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

Sửa lại là 1/256 nha

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

\(=1-\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{4}-\frac{1}{4}\right)-\left(\frac{1}{8}-\frac{1}{8}\right)-...-\left(\frac{1}{128}-\frac{1}{128}\right)-\frac{1}{256}\)

\(=1-\frac{1}{256}=\frac{255}{256}\)

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

\(=1-\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{4}-\frac{1}{4}\right)-\left(\frac{1}{8}-\frac{1}{8}\right)-...-\left(\frac{1}{128}-\frac{1}{128}\right)-\frac{1}{256}\)

\(=1-\frac{1}{256}=\frac{255}{256}\)

ta có : A=1/2+1/4+..+1/1024

=> A=1/2¹+1/2²+..+1/2¹⁰

=> A.2=(1/2¹+1/2²+..+1/2¹⁰).2

=> A.2=1+1/2¹+1/2²+..+1/2⁹

=> 2A-A=(1+1/2¹+1/2²+..+1/2⁹)-(1/2¹+1/2²+..+1/2¹⁰)

=> A=1-1/2¹⁰

\(\frac{2174}{1024}\)

giúp tôi với

225/143 nha ban

158692,1211

các bạn ơi giúp mình với , mình cần gấp lắm

Mình chỉ biết là ra 1/16 thôi

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

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

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

=>A = 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

=1-1/256

=255/256 

\(\frac{1023}{1024}\)