Question

[phuonganh do]

Tính

\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)

\(B=\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}+\frac{1}{972}\)

Answer 1

[Đồng Thiên Ái]

\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)

=>\(2A=1+\frac{1}{2}+...+\frac{1}{2^8}\)

=>\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)

                       \(=1-\frac{1}{2^9}=\frac{511}{512}\)

\(B=\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}+\frac{1}{972}\)

=>\(3B=\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}\)

=>\(3B-B=\left(\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+...+\frac{1}{324}\right)-\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+...+\frac{1}{972}\right)\)

=>\(2B=\frac{3}{4}-\frac{1}{972}=\frac{182}{243}\)

=>\(B=\frac{182}{243}:2=\frac{91}{243}\)