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

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

Lấy ( 2 ) - ( 1 ) ta được :

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

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

A = 1 + 3 + 3² + ... + 3¹⁰⁰

⇒ 3A = 3 + 3² + 3³ + ... + 3¹⁰¹

⇒ 2A = 3A - A 

= (3 + 3² + 3³ + ... + 3¹⁰¹) - (1 + 3 + 3² + ... + 3¹⁰⁰)

= 3¹⁰¹ - 1

⇒ A = (3¹⁰¹ - 1)/2

cảm ơn cậu

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

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

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

bai toan nay kho qua

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

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

3A=3+3^2+3^3+3^4+...+3^101

3A-A=(3+3^2+3^3+3^4+...+3^101)-(1+3+3^2+3^3+...+3^100)

2A=3^101-1

A=\(\frac{3^{101}-1}{2}\)

( Dấu . là dấu nhân đấy nha)

A = 1 + 3 + 3² + ... + 3¹⁰⁰

3A = 3 + 3² + 3³ + ... + 3¹⁰¹

3A - A = 2A = 3¹⁰¹ - 1
A = \(\frac{3^{101}-1}{2}\)

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

\(3A=3+3^2+........+3^{101}\)

\(3A-A=\left(3+3^2+.....+3^{101}\right)-\left(1+3+3^2+.....+3^{100}\right)\)

\(2A=3^{101}-1\)

\(A=\frac{3^{101}-3}{2}\)

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

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

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

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

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

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

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

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

\(2A=2+2^2+2^3+..,+2^{101}\)

\(2A-A=\left(2+2^2+2^3+...+2^{101}\right)-\left(1+2+2^2+...+2^{100}\right)\)

\(A=2^{101}-1\)

\(B=\)\(3-3^2+3^3-...-3^{100}\)

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

\(\Rightarrow3A=3+3^2+3^3+...+3^{101}\)

\(\Rightarrow3A-A=3^{101}-1\)

\(\Rightarrow A=\frac{3^{101}-1}{2}\)