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\(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+..........+\dfrac{1}{256}+\dfrac{1}{512}=?\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}+...+\dfrac{1}{256}-\dfrac{1}{512}-\dfrac{1}{512}\)
\(=1-\dfrac{1}{512}\)
\(=\dfrac{511}{512}\)
Vậy \(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+.........+\dfrac{1}{256}+\dfrac{1}{512}=\dfrac{511}{512}\)
Bài bạn trên cách trình bày mk ko hiểu lắm! mk làm lại nhé!
Đặt :
\(S=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...........+\dfrac{1}{256}+\dfrac{1}{512}\)
\(\Leftrightarrow S=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+.........+\dfrac{1}{2^8}+\dfrac{1}{2^9}\)
\(\Leftrightarrow2S=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+.........+\dfrac{1}{2^9}\right)\)
\(\Leftrightarrow2S=1+\dfrac{1}{2}+\dfrac{1}{2^2}+.........+\dfrac{1}{2^8}\)
\(\Leftrightarrow2S-S=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^8}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+......+\dfrac{1}{2^9}\right)\)
\(\Leftrightarrow S=1-\dfrac{1}{2^9}\)
\(\Leftrightarrow S=1-\dfrac{1}{512}=\dfrac{511}{512}\)
\(D=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+..........+\dfrac{1}{256}+\dfrac{1}{512}\)
\(\Leftrightarrow2D=1+\dfrac{1}{2}+\dfrac{1}{4}+......+\dfrac{1}{256}\)
\(\Leftrightarrow2D-D=\left(1+\dfrac{1}{2}+.....+\dfrac{1}{256}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+.....+\dfrac{1}{512}\right)\)
\(\Leftrightarrow D=1-\dfrac{1}{512}=\dfrac{511}{512}\)
\(2A=2\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
\(2A=1+\frac{1}{2}+...+\frac{1}{2^9}\)
\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
\(A=1-\frac{1}{2^{10}}=\frac{2^{10}-1}{2^{10}}=\frac{1023}{1024}\)
BẤM ĐÚNG NHÉ
Đặt \(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{256}+\dfrac{1}{512}\)
\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{128}+\dfrac{1}{256}\)
\(\Rightarrow A=2A-A=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{128}+\dfrac{1}{256}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-...-\dfrac{1}{256}-\dfrac{1}{512}\)
\(\Rightarrow A=1-\dfrac{1}{512}=\dfrac{511}{512}\)
Đặt A = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512
2A = 1/2 x 2 + 1/4 x 2 + 1/8 x 2 + 1/16 x 2 + 1/32 x 2 + 1/64 x 2 + 1/128 x 2 + 1/256 x 2 + 1/512 x 2
2A = 1 + 1/2 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256
2A - A = ( 1 + 1/2 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 ) - ( 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512 )
A = 1 - 1/512
A = 511/512
\(a,\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(=1-\frac{1}{7}\)
\(=\frac{6}{7}\)
\(b,\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
Ta có :
\(\frac{1}{2}=1-\frac{1}{2}\)
\(\frac{1}{4}=\frac{1}{2}-\frac{1}{4}\)
\(\frac{1}{8}=\frac{1}{4}-\frac{1}{8}\)
\(\frac{1}{16}=\frac{1}{8}-\frac{1}{16}\)
\(\frac{1}{32}=\frac{1}{16}-\frac{1}{32}\)
Thay vào ta có :
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}\)
\(=1-\frac{1}{32}\)
\(=\frac{31}{32}\)
\(c,\)\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\)
Ta có :
\(\frac{1}{2}=1-\frac{1}{2}\)
\(\frac{1}{4}=\frac{1}{2}-\frac{1}{4}\)
...................
\(\frac{1}{256}=\frac{1}{128}-\frac{1}{256}\)
Thay vào ta có :
\(=\)\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{128}-\frac{1}{256}\)
\(=1-\frac{1}{256}\)
\(=\frac{255}{256}\)
Đặt A=1/2+1/4+1/8+...+1/256
2A=2/4+2/8+2/16+...+2/512
2A—A=(2/4+2/8+2/16+...+2/512—1/2+1/4+1/8+...+1/256)
A=2/512—1/2
Q=\(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{256}\)
Q= \(1+\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+....+\left(\frac{1}{128}-\frac{1}{256}\right)\)
Q=\(1+1+\frac{1}{256}\)
Q=\(\frac{513}{256}\)
khác gì bài 1+1/2+1/2^2+...+1/2^8
nhân 2 rồi trừ đi là ra chứ có gì đâu mà nan giải hả bạn