Question

Tính :

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

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

(Các bn giúp mik vs mai mik phải nộp bài r )

Answer 1

\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\\ 2A=1+\dfrac{1}{2}+...+\dfrac{1}{2^8}\\ 2A-A=\left(1+\dfrac{1}{2}+...+\dfrac{1}{2^8}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)\\ A=1-\dfrac{1}{2^9}=\dfrac{511}{512}\)

\(B=\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}+\dfrac{1}{972}\\ 3B=\dfrac{3}{4}+\dfrac{3}{12}+\dfrac{3}{36}+\dfrac{3}{108}+\dfrac{3}{324}+\dfrac{3}{972}\\ 3B=\dfrac{3}{4}+\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}\\ 3B-B=\left(\dfrac{3}{4}+\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}\right)-\left(\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}+\dfrac{1}{972}\right)\\ 2B=\dfrac{3}{4}-\dfrac{1}{972}=\dfrac{182}{243}\\ B=\dfrac{364}{243}\)