\(\Rightarrow3.A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)

\(\Rightarrow3.A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\right)\)

\(2.A=1-\frac{1}{3^{100}}=\frac{3^{100}-1}{3^{100}}\Rightarrow A=\frac{3^{100}-1}{2.3^{100}}\)

\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\\ \Leftrightarrow3A=3\left(+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\right)\\ =1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)

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

Ta có: \(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\)

\(\Leftrightarrow3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)

\(\Leftrightarrow2\cdot A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}-\dfrac{1}{3}-\dfrac{1}{3^2}-...-\dfrac{1}{3^{100}}\)

\(\Leftrightarrow2\cdot A=1-\dfrac{1}{3^{100}}\)

\(\Leftrightarrow2\cdot A=\dfrac{3^{100}-1}{3^{100}}\)

\(\Leftrightarrow A=\dfrac{3^{100}-1}{2\cdot3^{100}}\)

bai toan nay kho qua

3A=1+1/3+1/3^2+...........+1/3^99

3A-A=1-1/3^100

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

A = 1/3 + 1/3^2 + 1/3^3 + ... + 1/3^100

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

A : 3 = 1/3^2 + 1/3^3 + ... + 1/3^101

A - A : 3 = 1/3 + 1/3^2 + ... + 1/3^100 - 1/3^2 - 1/3^3 - ... - 1/3^101

A . 2/3 = (1/3^2 - 1/3^2) + (1/3^3 - 1/3^3) + ... + (1/3^100 - 1/3^100) + ( 1/3 - 1/3^101 )

A . 2/3 = 0 + 0 + 0 + ... + 0 + 1/3 - 1/3^101

A . 2/3 = 1/3 - 1/3^101

=> A = 1/2 - 1/3^100.2

A = \(\dfrac{100-(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{100})}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{99}{100}}\)

Xét các mẫu số của dãy phân số : \(\dfrac{1}{1};\dfrac{1}{2};....;\dfrac{1}{100}\)

ta có dãy số: 1; 2; ....;100

Dãy số trên có số số hạng là: ( 100 - 1) : 1 + 1 = 100 (số)

Tách 100 thành tổng của 100 số 1 rồi nhóm lần lượt 1 với từng phân số thuộc dãy phân số trên khi đó ta có:

A = \(\dfrac{100-(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{100})}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+.....+\dfrac{99}{100}}\)

A = \(\dfrac{(1-1)+(1-\dfrac{1}{2})+(1-\dfrac{1}{3})+....+(1-\dfrac{1}{100})}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+.....+\dfrac{99}{100}}\)

A = \(\dfrac{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+....+\dfrac{99}{100}}\)

A = 1