Question

Mình đang cần gấp đáp án bài này, mong các bạn giúp đỡ:

Tính:

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

Cảm ơn mọi người! 😊😊😊

Answer 1

Đặt \(A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{20}}\)

\(\Rightarrow2A=2+1+\frac{1}{2}+...+\frac{1}{2^9}\)

\(\Rightarrow2A-A=\left(2+1+\frac{1}{2}+...+\frac{1}{2^9}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)

\(\Rightarrow A=2-\frac{1}{2^{10}}\)

Answer 2

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

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

\(2A-A=\left(2+1+\frac{1}{2}+...+\frac{1}{2^9}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)

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

Answer 3

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

Ta có : A =1 +  \(\frac{1}{2}\)+\(\frac{1}{2^2}\)...+\(\frac{1}{2^{10}}\)

=>     \(\frac{1}{2}\)A =  \(\frac{1}{2}\)+\(\frac{1}{2^2}\)...+\(\frac{1}{2^{10}}\)+ \(\frac{1}{2^{11}}\)

=> A - \(\frac{1}{2}\)A=    (     1 +  \(\frac{1}{2}\)+\(\frac{1}{2^2}\)...+\(\frac{1}{2^{10}}\) )   -  ( \(\frac{1}{2}\)+\(\frac{1}{2^2}\)...+\(\frac{1}{2^{10}}\)+ \(\frac{1}{2^{11}}\))

=>  \(\frac{1}{2}\)A = 1 -  \(\frac{1}{2^{11}}\)

=>  \(\frac{1}{2}\)A= \(\frac{2^{11}-1}{2^{11}}\)

=> A = \(\frac{2^{11}-1}{2^{10}}\)

Vậy A = \(\frac{2^{11}-1}{2^{10}}\)