Xin lỗi, nhìn nhầm:

A = 3^100 - 3^99 + 3^98 - 3^97 +...........+ 3^2 - 3 + 1 
3A = 3^101 - 3^100 + 3^99 - 3^98 +...+3^3 -3^2 +3 
=> 4A = 3A + A =  3^101 + 1 
A = \(\frac{3^{101}+1}{4}\)

B = 3^100 - 3^99 + 3^98 - 3^97 +...........+ 3^2 - 3 + 1 
3B = 3^101 - 3^100 + 3^99 - 3^98 +...+3^3 -3^2 +3 
Cộng vế với vế triệt tiêu, ta có : 
4B = 3^101 + 1 
B = \(\frac{3^{101}+1}{4}\)

Bài 1: 

a: \(2A=2^{101}+2^{100}+...+2^2+2\)

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

b: \(3B=3^{101}+3^{100}+...+3^2+3\)

\(\Leftrightarrow2B=3^{100}-1\)

hay \(B=\dfrac{3^{100}-1}{2}\)

c: \(4C=4^{101}+4^{100}+...+4^2+4\)

\(\Leftrightarrow3C=4^{101}-1\)

hay \(C=\dfrac{4^{101}-1}{3}\)

A = 2¹⁰⁰ - 2⁹⁹ + 2⁹⁸ - 2⁹⁷ + ...+ 2² - 2

2.A = 2¹⁰¹ - 2¹⁰⁰ + 2⁹⁹ - 2⁹⁸ + ...+ 2³ - 2²

A + 2.A =  2¹⁰¹ - 2 => 3.A = 2¹⁰¹ - 2 => A = (2¹⁰¹ - 1) / 3

B : tương tự

\(C=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\)

\(3C=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}+\frac{1}{3^{98}}\)

\(3C-C=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\right)\)

\(2C=1-\frac{1}{3^{99}}< 1\)

\(\Rightarrow C=\frac{1-\frac{1}{3^{99}}}{2}< \frac{1}{2}\)

1.

B = 3¹⁰⁰ - 3⁹⁹ + 3⁹⁸ - 3⁹⁷ + ... + 3² - 3 + 1

3B = 3¹⁰¹ - 3¹⁰⁰ + 3⁹⁹ - 3⁹⁸ + ... + 3³ - 3² + 3

3B + B = ( 3¹⁰¹ - 3¹⁰⁰ + 3⁹⁹ - 3⁹⁸ + ... + 3³ - 3² + 3 ) + ( 3¹⁰⁰ - 3⁹⁹ + 3⁹⁸ - 3⁹⁷ + ... + 3² - 3 + 1 )

4B = 3¹⁰¹ + 1

B = \(\frac{3^{101}+1}{4}\)

\(A=3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3^1+1\)
=) \(3A=3.\left(3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3+1\right)\)
= \(3^{101}-3^{100}+3^{99}-3^{98}+...+3^3-3^2+3^1\)
=) \(3A+A=3^{101}-3^{100}+3^{99}-3^{98}+...+3^3-3^2+3^1+3^{100}-3^{99}\)
+ \(3^{98}-3^{97}+...+3^2-3^1+1\)
=) \(4A=3^{101}+1\)
=) \(A=\frac{3^{101}+1}{4}\)

Dùng sai phân như sau

\(3Q=3^{101}-3^{100}+3^{99}-...-3^2+3\)

\(Q=3^{100}-3^{99}+3^{98}-...-3+1\)

Cộng 2 biểu thức trên theo vế,ta có:

\(4Q=3^{101}+1\Rightarrow Q=\frac{3^{101}+1}{4}\)