Question

Tính nhanh :

\(\frac{1}{3}\)+\(\frac{1}{9}\)+ \(\frac{1}{27}\)+ \(\frac{1}{81}\)+ \(\frac{1}{243}\)+ \(\frac{1}{729}\)+ \(\frac{1}{2187}\)

Giúp mình nha !

Answer 1

Đặt \(D=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+\frac{1}{2187}\)

\(\Leftrightarrow D=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}+\frac{1}{3^7}\)

\(\Leftrightarrow3D=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}\)

\(\Leftrightarrow3D-D=2D=1-\frac{1}{3^6}\)

\(\Leftrightarrow D=\left(1-\frac{1}{3^6}\right)\div2\)

Answer 2

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Answer 3

Đặt  \(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+\frac{1}{2187}\)

\(A=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)

\(\frac{1}{3}A=\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\)

\(A-\frac{1}{3}A=\left(\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\right)\)

\(\frac{2}{3}A=\frac{1}{3}-\frac{1}{3^8}\)

\(\Rightarrow A=\left(\frac{1}{3}-\frac{1}{3^8}\right):\frac{2}{3}\)

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Answer 4

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\(A=\left(\frac{1}{3}-\frac{1}{3^8}\right):\frac{2}{3}\)

\(A=\frac{2186}{6561}:\frac{2}{3}\)

\(A=\frac{1093}{2187}\)

Answer 5

   Đặt  \(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+\frac{1}{2187}\)

\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}+\frac{1}{3^7}\)

\(3A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}\)

\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}+\frac{1}{3^7}\right)\)

\(2A=1-\frac{1}{3^7}\)

\(A=\left(1-\frac{1}{3^7}\right)\text{ : }2\)