Question

Tính:\(S=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+...\)

Answer 1

\(S=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{2^n}=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^n}\)

=>\(\frac{S}{2}=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{n+1}}\)

=> \(\frac{S}{2}-S=\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+..+\frac{1}{2^{n+1}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+..+\frac{1}{2^n}\right)\)

=> \(-\frac{S}{2}=\frac{1}{2^{n+1}}-1\)

=> S= \(2-\frac{1}{2^n}\)