Question

A=\(\dfrac{1}{2}\)-\(\dfrac{1}{2^2}\)+\(\dfrac{1}{2^3}\)-\(\dfrac{1}{2^4}\)+...+\(\dfrac{1}{2^{299}}\)-\(\dfrac{1}{2^{100}}\)

Answer 1

\(A=\dfrac{1}{2}-\dfrac{1}{2^2}+\dfrac{1}{2^3}-\dfrac{1}{2^4}+...+\dfrac{1}{2^{99}}-\dfrac{1}{2^{100}}\)

\(\Rightarrow2A=1-\dfrac{1}{2}+\dfrac{1}{2^2}-\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}-\dfrac{1}{2^{99}}\)

\(\Rightarrow3A=A+2A\)

\(=\left(\dfrac{1}{2}-\dfrac{1}{2^2}+\dfrac{1}{2^3}-\dfrac{1}{2^4}+...+\dfrac{1}{2^{99}}-\dfrac{1}{2^{100}}\right)+\left(1-\dfrac{1}{2}+\dfrac{1}{2^2}-\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}-\dfrac{1}{2^{99}}\right)\)

\(=1-\dfrac{1}{2^{100}}\)

\(\Rightarrow A=\dfrac{1-\dfrac{1}{2^{100}}}{3}\)

Answer 2

Em kiểm tra lại đề xem đã ghi đề đúng chưa nhé, ở số hạng cuối cùng ấy