Question

Tính:

a)      \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{100}}\)

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Answer 1

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)

\(A=1-\frac{1}{2^{100}}\)

\(A=\frac{2^{100}-1}{2^{100}}\)

Answer 2

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)

\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+..+\frac{1}{2^{100}}\right)\)

\(A=1-\frac{1}{2^{100}}\)

hok tốt!!

Answer 3

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)

\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)

\(\Rightarrow A=1-\frac{1}{2^{100}}\)

Vậy \(A=1-\frac{1}{2^{100}}\)

Answer 4

\(A=\)nhưu  trên

=>\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)

\(=>A-2A=\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)\)

=>\(-A=\frac{1}{2^{100}}-1=\frac{1-2^{100}}{2^{100}}\)

=>\(A=\frac{2^{100}-1}{2^{100}}\left(đúng\right)ko\)

Answer 5

A=\(1-\frac{1}{2^{100}}\)

Answer 6

Ta có : \(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\)   (1)

          \(2.A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)(2)

Từ (1) và (2) => ta có :

\(2.A-A=1-\frac{1}{2^{100}}\Rightarrow A=1-\frac{1}{2^{100}}\)

        

 Vậy A = \(1-\frac{1}{2^{100}}\)