Question

Tính Nhanh:

B=\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{8}\)+\(\dfrac{1}{16}\)+\(\dfrac{1}{32}\)+\(\dfrac{1}{64}\)

Giúp mình với mình cần gấp ạ

Answer 1

\(B=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)

=>\(B=\dfrac{32}{64}+\dfrac{16}{64}+\dfrac{6}{64}+\dfrac{2}{64}+\dfrac{1}{64}\)

=>\(B=\dfrac{32+16+6+2+1}{64}\)

=>\(B=\dfrac{63}{64}\)

Answer 2

\(\dfrac{63}{64}\)

Answer 3

\(B=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^6}\)

\(2B=1+\dfrac{1}{2}+...+\dfrac{1}{2^5}\)

\(\Rightarrow2B-B=1-\dfrac{1}{2^6}\)

\(\Rightarrow B=1-\dfrac{1}{2^6}=1-\dfrac{1}{64}=\dfrac{63}{64}\)