Question

Tính tổng sau đây bằng cách thuận tiện nhất:

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

Giúp mk nha 

Answer 1

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

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

\(3A=3\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\right)\)

\(3A=3+1+...+\frac{1}{3^4}\)

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

\(2A=3-\frac{1}{3^5}\)

\(A=\frac{3-\frac{1}{3^5}}{2}\)

 

 

Answer 2

Đặt \(S=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)

      \(S=1+\frac{1}{1\times3}+\frac{1}{3\times3}+\frac{1}{9\times3}+\frac{1}{27\times3}+\frac{1}{81\times3}\)

\(S\times3=\left(1+\frac{1}{1\times3}+\frac{1}{3\times3}+\frac{1}{9\times3}+\frac{1}{27\times3}+\frac{1}{81\times3}\right)\times3\)

\(S\times3=3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)

Xét: \(S\times3-S=\left(3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\right)-\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)

              \(S\times2=3-\frac{1}{243}\)

              \(S\times2=\frac{728}{243}\)

                    \(S=\frac{728}{243}\div2\)

                    \(S=\frac{364}{243}\)

Vậy \(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}=\frac{364}{243}\)

Answer 3

khó vậy nhỉ

Answer 4

Đặt 1/3 ra