Gọi \(\dfrac{1}{3}\)+\(\dfrac{1}{3^2}\)+\(\dfrac{1}{3^3}\)+...+\(\dfrac{1}{3^8}\) là A
\(\Rightarrow\)A= \(\dfrac{1}{3}\)+\(\dfrac{1}{3^2}\)+\(\dfrac{1}{3^3}\)+...+\(\dfrac{1}{3^8}\)
\(\Rightarrow\)3A= 1+ \(\dfrac{1}{3}\)+\(\dfrac{1}{3^2}\)+\(\)...+\(\dfrac{1}{3^7}\)
\(\Rightarrow\)3A-A= (1+ \(\dfrac{1}{3}\)+\(\dfrac{1}{3^2}\)+\(\)...+\(\dfrac{1}{3^7}\)) - (\(\dfrac{1}{3}\)+\(\dfrac{1}{3^2}\) +\(\dfrac{1}{3^3}\)+...+\(\dfrac{1}{3^8}\) )
\(\Rightarrow\)2A= 1+ \(\dfrac{1}{3}\)+\(\dfrac{1}{3^2}\)+\(\)...+\(\dfrac{1}{3^7}\) - \(\dfrac{1}{3}\)-\(\dfrac{1}{3^2}\)-\(\dfrac{1}{3^3}\)-...-\(\dfrac{1}{3^8}\)
= 1-\(\dfrac{1}{3^8}\)
= \(\dfrac{6550}{6561}\)
\(\Rightarrow\)A= \(\dfrac{6550}{6561}\):2= \(\dfrac{6550}{6561}\).\(\dfrac{1}{2}\)=\(\dfrac{3280}{6561}\)
Chúc bạn học tốt
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