a) \(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)

\(2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)

\(\Rightarrow A+2A=2^{101}-2\)

  \(A\left(1+2\right)=2^{101}-2\)

  \(A.3=2^{101}-2\)

  \(A=\frac{2^{101}-2}{3}\)

b) \(B=3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3\)

\(3B=3^{101}-3^{100}+3^{99}-3^{98}+...+3^3-3^2\)

\(\Rightarrow B+3B=3^{101}-3\)

\(B\left(1+3\right)=3^{101}-3\)

\(4B=3^{101}-3\)

   \(B=\frac{3^{101}-3}{4}\)

Thanks bạn

a, \(A=...\)

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

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

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

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

b, tương tự a \(B=\frac{3^{101}+1}{4}\)

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

   = ( 2¹⁰⁰ + 2⁹⁸ + ... + 2² ) - ( 2⁹⁹ + 2⁹⁷ + ... + 2 )

   = ( 2¹⁰⁰ + 2⁹⁸ + ... + 2² ) - 2( 2⁹⁹ + 2⁹⁷ + ... + 2 ) + ( 2⁹⁹ + 2⁹⁷ + ... + 2 )

   = 2⁹⁹ + 2⁹⁷ + ... + 2 

=> 4A = 2¹⁰³ + 2⁹⁹ + ... + 2³

=> 3A = 2¹⁰³ - 2

=> A = \(\frac{2^{103}-2}{3}\)

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

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

b) B = 2¹⁰⁰ - 2⁹⁹ + 2⁹⁸ - 2⁹⁷ + ...+ 2² - 2

=> B x 2 = 2^{101 -} 2¹⁰⁰ + 2⁹⁹ -  2⁹⁸  + ...2³ - 2²

=> B x 2 + B = (2^{101 -} 2¹⁰⁰ + 2⁹⁹ -  2⁹⁸  + ...2³ - 2² ) + (2¹⁰⁰ - 2⁹⁹ + 2⁹⁸ - 2⁹⁷ + ...+ 2² - 2)

  <=>  B x 3 = 2¹⁰¹ - 2 = 2. ( 2⁹⁹ - 1)

=> B = \(\frac{2.\left(2^{99}-1\right)}{3}\)

Phần c) Làm tương tự Lấy C x 3 rồi + với C.

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 = \(\frac{3^{101}+1}{4}\)

3^100-(3^99+3^98+.............+3^1+1)

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 = 3¹⁰¹ + 1

4
 

\(A=2^{100}-2^{99}+2^{98}-2^{97}+....+2^2-2\)

\(2A=2^{101}-2^{100}+2^{99}-2^{98}+....+2^3-2^2\)

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

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

b) tương tự

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