Bài làm:
Ta có: \(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}+\frac{1}{6561}=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}+\frac{1}{3^8}\)
=> \(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}+\frac{1}{3^7}\)
=> \(3A-A=\left(1+\frac{1}{3}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\right)\)
<=> \(2A=1-\frac{1}{3^8}=\frac{3^8-1}{3^8}\)
=> \(A=\frac{3^8-1}{3^8.2}\)
Bài làm :
Ta có :
\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{6561}\)
\(\Rightarrow3\times A=\frac{1\times3}{3}+\frac{1\times3}{9}+\frac{1\times3}{27}+...+\frac{1\times3}{6561}\)
\(3\times A=1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{729}+\frac{1}{2187}\)
\(3\times A=1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{729}+\frac{1}{2187}+\left(\frac{1}{6561}-\frac{1}{6561}\right)\)
\(3\times A=1+\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{729}+\frac{1}{2187}+\frac{1}{6561}\right)-\frac{1}{6561}\)
\(3\times A=1+A-\frac{1}{6561}\)
\(\Rightarrow2\times A=1-\frac{1}{6561}\)( Trừ bỏ A ở cả 2 vế )
\(2\times A=\frac{6560}{6561}\)
\(A=\frac{6560}{6561}\div2=\frac{3280}{6561}\)
Vậy A=3280/6561
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