Question

\(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{6561}\)

Mong giúp đỡ ạ!!!!

Answer 1

Sửa đề: A=1/3+1/9+1/27+...+1/6561

=1/3+1/3^2+1/3^3+...+1/3^8

=>3A=1+1/3+...+1/3^7

=>3A-A=1-1/3^8

=>\(2A=\dfrac{3^8-1}{3^8}\)

=>\(A=\dfrac{3^8-1}{2\cdot3^8}\)

Answer 2

Đặt \(S=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{6561}\)

\(3S=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{2187}\)

\(2S=\dfrac{2188}{2187}-\left(\dfrac{1}{27}+\dfrac{1}{6561}\right)\)

\(2S=\dfrac{2188}{2187}-\dfrac{244}{6561}\)

\(2S=\dfrac{4376}{6561}-\dfrac{244}{6561}\)

\(2S=\dfrac{4132}{6561}\)

\(S=\dfrac{2066}{6561}\)