Question

\(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) +............ + \(\dfrac{1}{256}\) + \(\dfrac{1}{512}\) = ?

Answer 1

\(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+..........+\dfrac{1}{256}+\dfrac{1}{512}=?\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}+...+\dfrac{1}{256}-\dfrac{1}{512}-\dfrac{1}{512}\)

\(=1-\dfrac{1}{512}\)

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

Vậy \(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+.........+\dfrac{1}{256}+\dfrac{1}{512}=\dfrac{511}{512}\)

Answer 2

Bài bạn trên cách trình bày mk ko hiểu lắm! mk làm lại nhé!

Đặt :

\(S=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...........+\dfrac{1}{256}+\dfrac{1}{512}\)

\(\Leftrightarrow S=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+.........+\dfrac{1}{2^8}+\dfrac{1}{2^9}\)

\(\Leftrightarrow2S=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+.........+\dfrac{1}{2^9}\right)\)

\(\Leftrightarrow2S=1+\dfrac{1}{2}+\dfrac{1}{2^2}+.........+\dfrac{1}{2^8}\)

\(\Leftrightarrow2S-S=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^8}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+......+\dfrac{1}{2^9}\right)\)

\(\Leftrightarrow S=1-\dfrac{1}{2^9}\)

\(\Leftrightarrow S=1-\dfrac{1}{512}=\dfrac{511}{512}\)

Answer 3

Ta có : \(\dfrac{1}{2}\) = 1 - \(\dfrac{1}{2}\) ; \(\dfrac{1}{4}\) = \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) ; \(\dfrac{1}{8}\) = \(\dfrac{1}{4}\) - \(\dfrac{1}{8}\) ;............. ;

\(\dfrac{1}{512}\) = \(\dfrac{1}{256}\) - \(\dfrac{1}{512}\)

Vậy : \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + ................... + \(\dfrac{1}{256}\) + \(\dfrac{1}{512}\)

= 1 - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{8}\) + ............. + \(\dfrac{1}{256}\) - \(\dfrac{1}{512}\)

= 1 - \(\dfrac{1}{512}\)

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