\(K=1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{243}+\frac{1}{729}\)
\(3K=3+1+\frac{1}{3}+...+\frac{1}{81}+\frac{1}{243}\)
\(3K-K=3-\frac{1}{729}\)
\(2K=\frac{2186}{729}\)
\(K=\frac{1093}{729}\)
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