Question

Tính giá trị biểu thức: \(A=\dfrac14+\dfrac1{4^2}+\dfrac1{4^3}+\dfrac1{4^4}+...+\dfrac1{4^{2020}}+\dfrac1{4^{2021}}.\)

Answer 1

\(A=\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2021}}\)

=>\(4A=1+\dfrac{1}{4}+...+\dfrac{1}{4^{2020}}\)

=>\(4A-A=1+\dfrac{1}{4}+...+\dfrac{1}{4^{2020}}-\dfrac{1}{4}-\dfrac{1}{4^2}-...-\dfrac{1}{4^{2021}}\)
=>\(3A=1-\dfrac{1}{4^{2021}}=\dfrac{4^{2021}-1}{4^{2021}}\)

=>\(A=\dfrac{4^{2021}-1}{4^{2021}\cdot3}\)

Answer 2

\(A=\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2020}}+\dfrac{1}{4^{2021}}\left(1\right)\)

\(\Rightarrow4A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2019}}+\dfrac{1}{4^{2020}}\left(2\right)\)

Lấy \(\left(2\right)\) trừ \(\left(1\right)\) ta được:

\(4A-A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2019}}+\dfrac{1}{4^{2020}}-\dfrac{1}{4}-\dfrac{1}{4^2}-\dfrac{1}{4^3}-...-\dfrac{1}{4^{2020}}-\dfrac{1}{4^{2021}}\)

\(\Rightarrow3A=1-\dfrac{1}{4^{2021}}\)

\(\Rightarrow3A=\dfrac{4^{2021}-1}{4^{2021}}\)

\(\Rightarrow A=\dfrac{4^{2021}-1}{4^{2021}}.\dfrac{1}{3}\)

\(\Rightarrow A=\dfrac{4^{2021}-1}{3.4^{2021}}\)

Answer 3

\(A=\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+\dfrac{1}{4^4}+...+\dfrac{1}{4^{2020}}+\dfrac{1}{4^{2021}}\)
\(4A=\dfrac{4}{4}+\dfrac{4}{4^2}+\dfrac{4}{4^3}+\dfrac{4}{4^4}+...+\dfrac{4}{4^{2020}}+\dfrac{4}{4^{2021}}\)
\(4A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2019}}+\dfrac{1}{4^{2020}}\)
\(4A-A=\left(1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2020}}\right)-\left(\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+\dfrac{1}{4^4}+...+\dfrac{1}{4^{2021}}\right)\)
\(3A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2020}}-\dfrac{1}{4}-\dfrac{1}{4^2}-\dfrac{1}{4^3}-\dfrac{1}{4^4}-...-\dfrac{1}{4^{2021}}\)
\(3A=1-\dfrac{1}{4^{2021}}\)
\(A=\dfrac{4^{2021}-1}{4^{2021}.3}\)