Question

Tính D = \(\frac{1}{2}-\frac{1}{2^4}+\frac{1}{2^7}-\frac{1}{2^{10}}+...-\frac{1}{2^{58}}\)

Answer 1

\(\frac{1}{2^3}\)D= \(\frac{1}{2^4}-\frac{1}{2^7}+\frac{1}{2^{10}}-\frac{1}{2^{13}}+...+\frac{1}{2^{58}}-\frac{1}{2^{61}}\)

D+ \(\frac{1}{2^3}\)D=\(\frac{1}{2}-\frac{1}{2^4}+\frac{1}{2^4}+\frac{1}{2^7}-\frac{1}{2^7}-\frac{1}{2^{10}}+\frac{1}{2^{10}}+...-\frac{1}{2^{58}}+\frac{1}{2^{58}}-\frac{1}{2^{61}}\)

\(\frac{9}{8}\)D= \(\frac{1}{2}-\frac{1}{2^{61}}\)=> D= \(\frac{\frac{1}{2}-\frac{1}{2^{61}}}{\frac{9}{8}}\)