Question

[Hàn Phong Nhi]

tính

\(\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+...+\frac{127}{128}-6\)

Answer 1

[Xyz OLM]

Đặt A =\(\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+...+\frac{127}{128}-6\)

= \(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{8}\right)+...+\left(1-\frac{1}{128}\right)-6\)

= \(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{2^2}\right)+\left(1-\frac{1}{2^3}\right)+...+\left(1-\frac{1}{2^7}\right)-6\)(7 cặp số)

= \(1-\frac{1}{2}+1-\frac{1}{2^2}+1-\frac{1}{2^3}+...+1-\frac{1}{2^7}-6\)

= \(\left(1+1+1+...+1\right)-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^7}-6\)

= \(1.7-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\right)-6\)

= \(7-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\right)-6\)

= \(7-6-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\right)\)

= \(1-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\right)\)

=> 2A = \(2-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\right)\)

Lấy 2A - A = \(\left(2-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\right)\right)-\left(1-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\right)\right)\)

              A  = \(2-1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^6}-1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\)

                  = \(2-1-1+\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^6}\right)\)

                  = \(0+\left(\frac{1}{2}-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^3}+...+\frac{1}{2^6}-\frac{1}{2^6}+\frac{1}{2^7}\right)\)

                  = \(0+\frac{1}{2^7}\)

                  = \(\frac{1}{2^7}\)