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Đặt A = \(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-\)...\(-\frac{1}{1024}\)
A= \(\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-\frac{1}{2^4}-\)....\(-\frac{1}{2^{10}}\)
2A=\(\frac{1}{1}\)\(-\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-\)...\(-\frac{1}{2^9}\)
2A-A=(\(\frac{1}{1}\)\(-\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-\)...\(-\frac{1}{2^{10}}\)) \(-\)(\(\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-\frac{1}{2^4}-\)..\(-\frac{1}{2^9}\))
A=\(1+\frac{1}{2^{10}}\)
A= \(\frac{1025}{1024}\)
1/1024 câu này trên violimpic vòng 2 và mình làm đúng rồi
\(A=\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
\(2A=\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-..-\frac{1}{512}\)
\(2A-A=\left(\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-..-\frac{1}{512}\right)-\left(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\right)\)
\(A=\frac{1}{4}+\frac{1}{4}-\frac{1}{2}+\frac{1}{1024}\)
\(A=\frac{1}{1024}\)
\(B=\frac{1}{2}-\frac{1}{4}-...-\frac{1}{1024}\)
\(=-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)\)
\(=-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
Đặt \(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}=A\)
\(2A=1+\frac{1}{2}+...+\frac{1}{2^9}\)
\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
\(A=1-\frac{1}{2^{10}}\).Thay A vào ta đc: \(B=-\left(1-\frac{1}{2^{10}}\right)\)
\(B=-\left(1-\frac{1}{1024}\right)\)
\(B=-\frac{1023}{1024}\)
Ta có:
\(-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
đặt \(A=1+\frac{1}{2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
\(\frac{1}{2}A=\frac{1}{2}+\frac{1}{2^3}+....+\frac{1}{2^{11}}\)
\(A-\frac{1}{2}A=\frac{1}{2}A\Rightarrow A=\frac{1-\frac{1}{2^{11}}}{\frac{1}{2}}=2-\frac{1}{2^{10}}\)
\(-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
\(=-1-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
\(A=1-\frac{1}{1024}=\frac{1023}{1024}\)
Vậy, \(-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}=-1-A=-1-\frac{1023}{1024}=-\frac{2047}{1024}\)
Đặt \(A=1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
\(\Rightarrow A=1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
\(2A=2-1-\frac{1}{2}-\frac{1}{4}-...-\frac{1}{512}\)
\(2A+A=\left(2-1-\frac{1}{2}-\frac{1}{4}-...-\frac{1}{512}\right)+\left(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\right)\)
\(\Rightarrow3A=2-\frac{1}{1024}\)
\(\Rightarrow3A=\frac{2048}{1024}-\frac{1}{1024}\)
\(\Rightarrow3A=\frac{2047}{1024}\)
\(\Rightarrow A=\frac{2047}{1024}:3\)
\(\Rightarrow A=\frac{2047}{3072}\)
gọi A=\(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\)
2xA=1+\(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)
2xA‐A=﴾1+\(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)﴿‐﴾\(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\)﴿
A=1‐\(\frac{1}{1024}\)
= \(\frac{1023}{1024}\)
vậy A=\(\frac{1023}{1024}\)
\(\text{Ta có: }\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-....-\frac{1}{1024}\)
\(=\left(1-\frac{1}{2}\right)-\left(\frac{1}{2}-\frac{1}{4}\right)-\left(\frac{1}{4}-\frac{1}{8}\right)-......-\left(\frac{1}{512}-\frac{1}{1024}\right)\)
\(=1-\frac{1}{1024}\)
\(=\frac{1023}{1024}\)
\(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-....-\frac{1}{1024}\)
\(=1-\left(1-\frac{1}{2}\right)-\left(\frac{1}{2}-\frac{1}{4}\right)-.......-\left(\frac{1}{512}-\frac{1}{1024}\right)\)
\(=1-1+\frac{1}{2}-\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+....+\frac{1}{512}-\frac{1}{1024}\)
\(=-\frac{1}{1024}\)
\(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-....-\frac{1}{1024}=1-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
=> \(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
=> \(A=2A-A=1-\frac{1}{2^{10}}\)
=> \(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-....-\frac{1}{1024}=1-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
\(=1-A=1-\left(1-\frac{1}{2^{10}}\right)=1-1+\frac{1}{2^{10}}\)
\(=\frac{1}{2^{10}}\)
ta có : \(\frac{1}{2}=1-\frac{1}{2};\frac{1}{4}=\frac{1}{2}-\frac{1}{4};\frac{1}{8}=\frac{1}{4}-\frac{1}{8}\)
\(\frac{1}{16}=\frac{1}{8}-\frac{1}{16};\frac{1}{1024}=\frac{1}{512}-\frac{1}{1024}\)
\(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-.....-\frac{1}{1024}\)
\(=1-\frac{1}{2}-\frac{1}{2}-\frac{1}{4}-\frac{1}{4}-\frac{1}{8}-\frac{1}{8}-\frac{1}{16}-\frac{1}{16}-....-\frac{1}{512}-\frac{1}{1024}\)
\(=1-\frac{1}{1024}\)
\(=\frac{1023}{1024}\)
đặt D=-(1/2+1/4+1/8+....+1/1024)
D=-(1/2+1/2^2+....+1/2^10)
đặt A=1/2+...+1/2^10
2A = 1+1/2+...+1/2^9
2A-A=(1+1/2+...+1/2^9)-(1/2+...+1/2^10)
A=1-1/2^10
A=2^10-1/2^10
D=-2^10-1/2^10