Question

Giúp mình giải câu hỏi này với:

1\2+1\4+1\8+...+1\1024

Answer 1

A=\(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{1024}\)

A=\(\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\)

=>2A=2.(\(\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\))

=>2A=\(1+\dfrac{1}{2}+...+\dfrac{1}{2^9}\)

=>2A-A=(\(1+\dfrac{1}{2}+...+\dfrac{1}{2^9}\))-(\(\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\))

=>A=1-\(\dfrac{1}{2^{10}}\)

Vậy \(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{1024}\)=1-\(\dfrac{1}{2^{10}}\)

Answer 2

\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\\ \Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\\ 2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\right)\\ A=1-\dfrac{1}{2^{10}}\)

Answer 3

Đặt \(S=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+....+\dfrac{1}{1024}\)

\(\Rightarrow S=\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+....+\dfrac{1}{2^{1024}}\)

\(\Rightarrow\dfrac{1}{2}S=\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+....+\dfrac{1}{2^{1025}}\)

\(\Rightarrow\dfrac{1}{2}S-S=\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+....+\dfrac{1}{2^{1024}}\right)-\left(\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+....+\dfrac{1}{2^{1024}}\right)\)

\(\Rightarrow\dfrac{1}{2}S=\dfrac{1}{2^{2014}}-\dfrac{1}{2}\)

\(\Rightarrow S=\left(\dfrac{1}{2^{2014}}-\dfrac{1}{2}\right).2=\dfrac{1}{2^{2013}}\)