Ta thấy:
\(P=\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+...+\frac{1}{4374}\\ =\frac{1}{2}\left(1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{2187}\right)\\ =\frac{1}{2}\left(\frac{1}{3^0}+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)\)
Mà:
\(\frac{1}{3}P=\frac{1}{2}\cdot\frac{1}{3}\left(\frac{1}{3^0}+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^7}\right)\\ =\frac{1}{2}\left(\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\right)\)
Suy ra: \(P-\frac{1}{3}P=\frac{1}{2}\left[\left(\frac{1}{3^0}+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\right)\right]\)
hay \(\frac{2}{3}P=\frac{1}{2}\left(\frac{1}{3^0}-\frac{1}{3^8}\right)=\frac{1}{2}\left(1-\frac{1}{6561}\right)=\frac{3280}{6561}\)
Vậy \(P=\frac{3280}{6561}:\frac{2}{3}=\frac{1640}{2187}\).
Chúc bạn học tốt nha
.
