Question

tinh A :biet A=1/2+1/2mu3+1/2mu5+.........+1/2mu99

Answer 1

A = \(\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+....+\frac{1}{2^{99}}\)

2²A = \(2+\frac{1}{2}+\frac{1}{2^3}+.....+\frac{1}{2^{97}}\)

=> 3A = \(2-\frac{1}{2^{99}}\)

=> A = \(\frac{2-\frac{1}{2^{99}}}{3}\)