Question

Tính tổng A=\(\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{19}}\)

Giúp mk nhanh nhé "lười suy nghĩ nên đành nhờ các bn"

Answer 1

\(A=\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{19}}\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{18}}\Rightarrow2A-A=A=1-\frac{1}{2^{19}}\)

lm hơi tắt nhỉ :D

Answer 2

\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{19}}\)

\(2A=1+\dfrac{1}{2}+...+\dfrac{1}{2^{18}}\)

\(2A-A=\left(1+\dfrac{1}{2}+...+\dfrac{1}{2^{18}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{19}}\right)\)

\(A=1-\dfrac{1}{2^{19}}\)