Question

\(\frac{1}{14}+\frac{1}{28}+\frac{1}{56}+...+\frac{1}{896}=x-3\)

Answer 1

Đặt:  \(A=\frac{1}{14}+\frac{1}{28}+\frac{1}{56}+....+\frac{1}{896}\)

\(\Rightarrow\)\(2A=\frac{1}{7}+\frac{1}{14}+\frac{1}{28}+....+\frac{1}{448}\)

\(\Rightarrow\)\(2A-A=\left(\frac{1}{7}+\frac{1}{14}+....+\frac{1}{448}\right)-\left(\frac{1}{14}+\frac{1}{28}+.....+\frac{1}{896}\right)\)

\(\Rightarrow\)\(A=\frac{1}{7}-\frac{1}{896}=\frac{127}{896}\)

mà  \(A=x-3\)

nên   \(x-3=\frac{127}{896}\)

\(\Rightarrow\)\(x=3\frac{127}{896}\)

Answer 2

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