\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\left(1\right)\)

\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\left(2\right)\)

          Lấy (2) - (1) ta được:\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+..+\frac{1}{3^{100}}\right)\)

                               \(\Leftrightarrow2A=1-\frac{1}{3^{100}}\)

                               \(\Leftrightarrow A=\left(\frac{3^{100}-1}{3^{100}}\right):2\)

                               \(\Leftrightarrow A=\frac{3^{100}-1}{2.3^{100}}\)

Cách làm là là A × 3 lên

K cho mk nha mn

Bạn phải giải thích cụ thể ra cho mk biết chứ

A× 3 = 1+ 1/3 + 1/3^2 + ...+ 1/3^99

A×3-a = 1- 1/ 3^100

A= (1/3^100): 2

\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)(1)

\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)(2)

Lấy (2) trừ đi (1) ta có :

\(2A=1-\frac{1}{3^{100}}\)

\(\Rightarrow A=\frac{\left(1-\frac{1}{3^{100}}\right)}{2}\)

ta có 3A=\(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)

\(\Rightarrow3A-A=\left(1+\frac{1}{3}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\right)\)

\(\Rightarrow2A=1-\frac{1}{3^{100}}\Rightarrow A=\frac{1-\frac{1}{3^{100}}}{2}\)

Đăt A = \(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+......+\frac{1}{7^{100}}\)

\(\Rightarrow7A=1+\frac{1}{7}+\frac{1}{7^2}+.....+\frac{1}{7^{100}}\)

\(\Rightarrow7A-A=1-\frac{1}{7^{100}}\)

\(\Rightarrow6A=1-\frac{1}{7^{100}}\)

\(\Rightarrow A=\frac{1-\frac{1}{7^{100}}}{6}\)

3A = 1 + 1/3 + 1/3^2 + ... + 1/3^199

3A - A = ( 1 + 1/3 + 1/3^2 + ... + 1/3^99 ) - ( 1/3 + 1/3^2 + 1/3^3 + ... + 1/3^100 )

2A = 1 - 1/3^100

A = ( 1 - 1/3^100 ) / 2

\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)

\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)

\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\right)\)

\(2A=1-\frac{1}{3^{100}}\)

\(A=\frac{3^{100}-1}{3^{100}.2}\)

mk chỉ làm được đến đây thôi

\(A=3+\frac{3}{1+2}+\frac{3}{1+2+3}+.....+\frac{3}{1+2+...+100}\)

     \(=3+\frac{3}{3}+\frac{3}{6}+...+\frac{3}{5050}\)

        \(=\frac{2}{2}.\left(3+\frac{3}{3}+\frac{3}{6}+...+\frac{3}{5050}\right)\)

          \(=\frac{6}{2}+\frac{6}{6}+\frac{6}{12}+...+\frac{6}{10100}\)

          \(=6.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\right)\)

            \(=6.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\right)\)

             \(=6.\left(1-\frac{1}{101}\right)\)

               \(=6.\frac{100}{101}=\frac{600}{101}\)

Vậy \(A=\frac{600}{101}\)

\(A=3+\frac{3}{1+2}+\frac{3}{1+2+3}+...+\frac{3}{1+2+...+100}\)

\(A=\frac{3.2}{2}+\frac{3.2}{\left(1+2\right).2}+\frac{3.2}{\left(1+2+3\right).2}+...+\frac{3.2}{\left(1+2+...+100\right).2}\)

\(A=\frac{6}{2}+\frac{6}{6}+\frac{6}{12}+...+\frac{6}{10100}\)

\(A=\frac{6}{1.2}+\frac{6}{2.3}+\frac{6}{3.4}+...+\frac{6}{100.101}\)

\(A=6\cdot\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\right)\)

\(A=6\cdot\left(1-\frac{1}{101}\right)=6\cdot\frac{100}{101}=\frac{600}{101}\)

Vay A = ........ 

A = 1 + 1/2 + 1/3 + 1/4 + 1/5 + ... + 1/100

Ta đổi A = 2-1+1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/99-1/100

A= 2 - 1 - 1/100 =200/100 -100/100 - 1/100

A= 99/100

Cảm ơn bạn Kudo Shinichi, nhưng 

1=2-1 ->ok

1/2=1-1/2 ->ok

1/3=1/2-1/3 -> sai 

vì 1/2-1/3=1/6