Câu hỏi của Nguyễn Nhật Giang

Question

Tính hơp lý : A = 1/3+1/9+1/27+1/81+1/243

Dang Tung · Accepted Answer

$A=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\
=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+\dfrac{1}{3^5}\
=>3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}\
=>3A-A=2A=1-\dfrac{1}{3^5}\
=>A=\dfrac{1-\dfrac{1}{3^5}}{2}=\dfrac{3^5-1}{2.3^5}$Câu trả lời: $A=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\
=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+\dfrac{1}{3^5}\
=>3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}\
=>3A-A=2A=1-\dfrac{1}{3^5}\
=>A=\dfrac{1-\dfrac{1}{3^5}}{2}=\dfrac{3^5-1}{2.3^5}$

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