Bài 5:
\(A=\left(1-\dfrac{1}{5}\right)\left(1-\dfrac{1}{6}\right)\cdot...\cdot\left(1-\dfrac{1}{20}\right)\)
\(=\dfrac{4}{5}\cdot\dfrac{5}{6}\cdot...\cdot\dfrac{19}{20}\)
\(=\dfrac{4}{20}=\dfrac{1}{5}\)
Bài 4:
a: \(A=1+\dfrac{1}{2}+...+\dfrac{1}{2048}\)
=>\(A=1+\dfrac{1}{2}+...+\left(\dfrac{1}{2}\right)^{11}\)
=>\(2A=2+1+...+\left(\dfrac{1}{2}\right)^{10}\)
=>\(2A-A=2+1+...+\left(\dfrac{1}{2}\right)^{10}-1-\dfrac{1}{2}-...-\left(\dfrac{1}{2}\right)^{11}\)
=>\(A=2-\dfrac{1}{2^{11}}=\dfrac{2^{12}-1}{2^{11}}\)
b: \(B=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{10}}\)
=>\(3B=1+\dfrac{1}{3}+...+\dfrac{1}{3^9}\)
=>\(3B-B=1+\dfrac{1}{3}+...+\dfrac{1}{3^9}-\dfrac{1}{3}-...-\dfrac{1}{3^{10}}\)
=>\(2B=1-\dfrac{1}{3^{10}}=\dfrac{3^{10}-1}{3^{10}}\)
=>\(B=\dfrac{3^{10}-1}{2\cdot3^{10}}\)
