Question

F=\(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+......... +\dfrac{1}{2^{2018}}\).Tính F

Answer 1

\(F=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2018}}\)

\(\dfrac{1}{2}F=\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{2019}}\)

\(F-\dfrac{1}{2}F=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2018}}-\dfrac{1}{2^2}-\dfrac{1}{2^3}-\dfrac{1}{2^4}-...-\dfrac{1}{2^{2019}}\)

\(\dfrac{1}{2}F=\dfrac{1}{2}-\dfrac{1}{2^{2019}}\)

\(F=\left(\dfrac{1}{2}-\dfrac{1}{2^{2019}}\right):\dfrac{1}{2}\)

\(F=\left(\dfrac{1}{2}-\dfrac{1}{2^{2019}}\right).2\)

\(F=1-\dfrac{1}{2^{2018}}\)

Answer 2

Cách khác

Answer 3

đậu phộng rang nó

gửi toàn câu mất dạy

Answer 4

coi chừng tao kêu công an