Question

[nguyenhamy]

chứng minh D=1/2+(1/2)^2+(1/2)^3+....+(1/2)^99<1

Answer 1

[Nguyễn Đình Dũng]

Ta có : 

D = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)

=> 2D = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)

=> 2D - D = \(1-\frac{1}{2^{98}}\)

=> D = \(1-\frac{1}{2^{98}}\)\(<1\)