Question

[Shibuki Ran]

Tinhs tổng sau:
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{2048}\)

(Nhập kết quả là phân số tói giản)

Answer 1

[DanAlex]

Đặt: \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2048}\)

\(\Rightarrow A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{11}}\)

\(\Rightarrow2A=2\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{11}}\right)\)

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

\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{11}}\right)\)

\(\Rightarrow A=1-\frac{1}{2^{11}}=\frac{2^{11}-1}{2^{11}}=\frac{2047}{2048}\)
 

Answer 2

[Justin Bieber]

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

\(2A-A=\left(1+...+\frac{1}{1024}\right)-\left(\frac{1}{2}+...+\frac{1}{2048}\right)\)

\(A=1-\frac{1}{2048}=\frac{2047}{2048}\)