Question

Tính giá trị của biểu thức A:

\(A=-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)

Answer 1

A= -(1+1/2+1/4+1/8+...+1/1024)

A=-(1+1/2+1/2^2+1.2^3+...1/2^10)

2A= -(2+1+1/2+1/^2+...1/2^9)

A=2A-A = -(2+1+1/2+1/^2+...1/2^9)-(1+1/2+1/2^2+1.2^3+...1/2^10) = -(2+1/2^10) = -2-1/2^10= -(2049/1024)

Answer 2

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

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

\(\Rightarrow-2A=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2056}\)

\(\Rightarrow-2A-A=\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2056}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)\)

\(\Rightarrow-3A=\frac{1}{2056}-1\)

\(\Rightarrow-3A=\frac{-2055}{2056}\)

\(\Rightarrow A=\frac{685}{2056}\)

Vậy...

Answer 3

\(\Rightarrow A=-\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)

Đặt B=\(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\)\(\Rightarrow2B=2+1+\frac{1}{2}+...+\frac{1}{512}\)

\(\Rightarrow2B-B=\)\(\left(2+1+\frac{1}{2}+...+\frac{1}{512}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)\)

\(\Rightarrow B=2-\frac{1}{1024}\)=\(\frac{2047}{1024}\)

=>A=-2047/1024

Answer 4

Sửa lại:

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

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

\(\Rightarrow-2A=2+1+\frac{1}{2}+...+\frac{1}{512}\)

\(\Rightarrow-2A-A=\left(2+1+\frac{1}{2}+...+\frac{1}{512}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)\)

\(\Rightarrow-3A=2-\frac{1}{1024}\)

\(\Rightarrow-3A=\frac{2047}{1024}\)

\(\Rightarrow A=\frac{-2048}{3}\)

Vậy...

Answer 5

\(A=-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)

\(A=-\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)

\(A=-\left(\frac{1}{2^0}+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)

Đặt \(B=\frac{1}{2^0}+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)

\(\frac{1}{2}B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{11}}\)

\(B-\frac{1}{2}B=\frac{1}{2^0}-\frac{1}{2^{11}}\)

\(\frac{1}{2}B=\frac{1}{2^0}-\frac{1}{2^{11}}\)

\(B=2\left(\frac{1}{2^0}-\frac{1}{2^{11}}\right)\)

Thế B vào A ta có

\(A=-2\left(\frac{1}{2^0}-\frac{1}{2^{11}}\right)\)