Đặt tổng là A

\(2xA=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{64}+\dfrac{1}{128}\)

\(\Rightarrow A=2xA-A=1-\dfrac{1}{256}=\dfrac{255}{256}\)

Đặt: \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}+\frac{1}{256}\)

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

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\)

\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^2}\right)-\left(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\right)\)

\(\Rightarrow A=1-\frac{1}{2^7}\)

E= 1/2 + 1/4 + 1/8 + 1/16 + ... + 1/128 + 1/256

2E = 2 ( 1/2 + 1/4 + 1/8 + 1/16 + ... + 1/128 + 1/256 )

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

=> E = 2E - E

= (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

k nhá, thanks

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

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

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

\(2E-E=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\right)\)

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

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

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

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

2S= 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 

=>S = 2S-S =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

255/256

ht

và

     NHỚ K CHO MIK NHA!!!!!!!!!!!!!!!!!

      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   = \(\dfrac{255}{256}\)

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

\(Ax2=2x\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}+\frac{1}{256}\right)\)

\(Ax2=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\)

\(Ax2-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\right)-\left(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{128}-\frac{1}{256}\right)\)

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

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

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

\(2\times E=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{128}\)

\(2\times E-E=\left(1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{128}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{256}\right)\)

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

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

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

=255/256

~hok tốt~

#Trang#

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

Dùng sai phân như sau

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

Vậy \(S=1-\frac{1}{256}\)

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

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

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

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

#Hok tốt