\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{512}-\frac{1}{1024}\)

\(=1-\frac{1}{1024}\)

\(=\frac{1023}{1024}\)

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}.\)

Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)

<=> \(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}+\frac{1}{512}\)

<=> \(2A-A=1+\frac{1}{2}+\frac{1}{4}+....+\frac{1}{256}+\frac{1}{512}-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{512}-\frac{1}{1024}\)

<=> \(A=1-\frac{1}{1024}\)

<=> \(A=\frac{1023}{1024}\)

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

=>\(2A=-2-1-\frac{1}{2}-\frac{1}{4}-...-\frac{1}{512}\)

=>\(2A-A=\left(-2-1-\frac{1}{2}-\frac{1}{4}-...-\frac{1}{512}\right)-\left(-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\right)\)

=>\(A=-2+\frac{1}{1024}\)

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

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

\(\Leftrightarrow-\dfrac{1}{2}A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{11}}\)

\(\Leftrightarrow A\cdot\dfrac{1}{2}=\dfrac{1}{2^{11}}-1\)

hay \(A=\dfrac{2\cdot\left(1-2^{11}\right)}{2^{11}}=\dfrac{1-2^{11}}{2^{10}}\)

Ta có:

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

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

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

\(=-1+\frac{-1023}{1024}\)

\(=-\frac{2047}{1024}\)

\(=\frac{1023}{1024}\)nha bn

1+1/2/+1/4+1/8+...+1/1024

=1+(1-1/2)+(1/2-1/4)+(1/4-1/8)+...(1/512-1/1024)

=1+1-1/2+1/2-1/4+1/4-1/8+...+1/512-1/1024

=1+1-1/1024

=2-1/1024

=2047/1024

kết quả là 1 số thuộc R

gọi A=1/2+1/4+1/8+...+1/1024

2xA=1+1/2+1/4+.....+1/512

2xA-A=(1+1/2+1/4+....+1/512)-(1/2+1/4+1/8+...+1/1024)

A=1-1/1024

=1023/1024

vậy A=1023/1024

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= \(\frac{1025}{1024}\)