Question

[Hatsune Miku]

1/2+1/2.2+1/2.2.2+1/2.2..... 50 thừa số

Answer 1

[Xyz OLM]

Đặt A = \(\frac{1}{2}+\frac{1}{2.2}+\frac{1}{2.2.2}+...+\frac{1}{2.2.2.....2}\)

          = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{50}}\)

=> 2A = \(2.\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{50}}\right)\)

           = \(2\times\frac{1}{2}+2\times\frac{1}{2^2}+2\times\frac{1}{2^3}+...+2\times\frac{1}{2^{50}}\)

           = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{49}}\)

 Lấy 2A - A = \(\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{49}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{50}}\right)\)

               A  = \(1+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{50}}\)

                   = \(1-\frac{1}{2^{50}}\)

Vậy  \(\frac{1}{2}+\frac{1}{2.2}+\frac{1}{2.2.2}+...+\frac{1}{2.2.2.....2}\)=  \(1-\frac{1}{2^{50}}\)