Question

Giúp mình với 

Answer 1

Đặt \(A=\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2048}\)

=>\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{11}}\)

=>\(2A=1+\dfrac{1}{2}+...+\dfrac{1}{2^{10}}\)

=>\(2A-A=1+\dfrac{1}{2}+...+\dfrac{1}{2^{10}}-\dfrac{1}{2}-\dfrac{1}{2^2}-...-\dfrac{1}{2^{11}}\)

=>\(A=1-\dfrac{1}{2^{11}}=\dfrac{2^{11}-1}{2^{11}}\)

\(\dfrac{1}{3}-\dfrac{1}{6}-\dfrac{1}{12}-...-\dfrac{1}{6144}\)

\(=\dfrac{1}{3}-\dfrac{1}{3}\cdot A\)

\(=\dfrac{1}{3}-\dfrac{1}{3}\cdot\dfrac{2^{11}-1}{2^{11}}=\dfrac{2^{11}-2^{11}+1}{3\cdot2^{11}}=\dfrac{1}{3\cdot2^{11}}\)

Answer 2

Đặt \(A=\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2048}\)

=>\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{11}}\)

=>\(2A=1+\dfrac{1}{2}+...+\dfrac{1}{2^{10}}\)

=>\(2A-A=1+\dfrac{1}{2}+...+\dfrac{1}{2^{10}}-\dfrac{1}{2}-\dfrac{1}{2^2}-...-\dfrac{1}{2^{11}}\)

=>\(A=1-\dfrac{1}{2^{11}}=\dfrac{2^{11}-1}{2^{11}}\)

\(\dfrac{1}{3}-\dfrac{1}{6}-\dfrac{1}{12}-...-\dfrac{1}{6144}\)

\(=\dfrac{1}{3}-\dfrac{1}{3}\cdot A\)

\(=\dfrac{1}{3}-\dfrac{1}{3}\cdot\dfrac{2^{11}-1}{2^{11}}=\dfrac{2^{11}-2^{11}+1}{3\cdot2^{11}}=\dfrac{1}{3\cdot2^{11}}\)