Question

1+1/3+1/9+1/27+....+1/2187

Answer 1

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Answer 2

Đặt \(B=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}\)

\(\Rightarrow3B=3.\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}\right)\)

\(\Rightarrow3B=3+1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{729}\)

\(\Rightarrow3B-B=\left(3+1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{729}\right)-\)\(\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}\right)\)

\(\Rightarrow2B=3-\frac{1}{2187}\)

\(\Rightarrow B=\left(3-\frac{1}{2187}\right):2\)

\(\Rightarrow B=\frac{6560}{2187}\)

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Answer 3

Đặt A = \(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}\)

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

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

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

            2A   = \(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}-1-\frac{1}{3}-\frac{1}{3^2}-\frac{1}{3^3}-...-\frac{1}{3^7}\)

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

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

\(\Rightarrow1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}=\left(3-\frac{1}{3^7}\right):2\)